Trigonometrical Identities

Solving,

⇒ sin⁴ θ - cos⁴ θ

⇒ (sin² θ + cos² θ)(sin² θ - cos² θ)

Substituting, sin² θ + cos² θ = 1, we get :

⇒ 1 × (sin² θ - cos² θ)

⇒ sin² θ - cos² θ

⇒ sin² θ - (1 - sin² θ)

⇒ sin² θ + sin² θ - 1

⇒ 2 sin² θ - 1.

Hence, Option 3 is the correct option.

Solving,

⇒ (1 + tan θ)² + (1 - tan θ)²

⇒ 1 + tan² θ + 2 tan θ + 1 + tan² θ - 2 tan θ

⇒ 2 + 2 tan² θ

⇒ 2(1 + tan² θ)

Substituting, 1 + tan² θ = sec² θ, we get :

⇒ 2 sec² θ.

Hence, Option 2 is the correct option.

Solving,

⇒ cot⁴ θ + cot² θ

⇒ cot² θ(cot² θ + 1)

⇒ cot² θ.cosec² θ

Substituting, cot² θ = cosec² θ - 1, we get :

⇒ (cosec² θ - 1).cosec² θ

⇒ cosec⁴ θ - cosec² θ.

Hence, Option 4 is the correct option.

Solving,

⇒ $\frac{1}{\text{1 - sin A}}$

Multiplying numerator and denominator by (1 + sin A), we get :

$$\begin{aligned}\Rightarrow \frac{1}{\text{1 - sin A}} \times \frac{\text{1 + sin A}}{\text{1 + sin A}} \\[1em] \Rightarrow \frac{\text{1 + sin A}}{1 - \text{sin}^2 A} \\[1em] \Rightarrow \frac{\text{1 + sin A}}{\text{cos}^2 A} \\[1em] \Rightarrow \frac{1}{\text{cos}^2 A} + \frac{\text{sin A}}{\text{cos}^2 A} \\[1em] \Rightarrow \frac{1}{\text{cos}^2 A} + \frac{1}{\text{cos}^2 A}. \text{ sin A} \\[1em] \Rightarrow \text{sec}^2 A + \text{sec}^2 A.\text{ sin A} \\[1em] \Rightarrow \text{sec}^2 A\text{(1 + sin A)}.\end{aligned}$$

Hence, Option 2 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{\text{tan}^2 A}{\text{(sec A + 1)}^2} \\[1em] \Rightarrow \frac{\text{sec}^2 A - 1}{\text{(sec A + 1)}^2} \\[1em] \Rightarrow \frac{\text{(sec A + 1)(sec A - 1)}}{\text{(sec A + 1)}^2} \\[1em] \Rightarrow \frac{\text{sec A - 1}}{\text{sec A + 1}} \\[1em] \Rightarrow \frac{\frac{1}{\text{cos A}} - 1}{\frac{1}{\text{cos A}} + 1} \\[1em] \Rightarrow \frac{\frac{\text{1 - cos A}}{\text{cos A}}}{\frac{\text{1 + cos A}}{\text{cos A}}} \\[1em] \Rightarrow \frac{\text{1 - cos A}}{\text{1 + cos A}}.\end{aligned}$$

Hence, Option 2 is the correct option.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{1}{\text{tan A + cot A}} \\[1em] \Rightarrow \frac{1}{\frac{\text{sin A}}{\text{cos A}} + \frac{\text{cos A}}{\text{sin A}}} \\[1em] \Rightarrow \frac{1}{\frac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}}} \\[1em] \Rightarrow \frac{\text{sin A cos A}}{\text{sin}^2 A + \text{cos}^2 A}\end{aligned}$$

By formula,

sin² A + cos² A = 1

$\Rightarrow \text{sin A cos A}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{1}{\text{tan A + cot A}} = \text{cos A sin A}$.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \text{tan A - cot A} \\[1em] \Rightarrow \frac{\text{sin A}}{\text{cos A}} - \frac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \frac{\text{sin}^2 A - \text{cos}^2 A}{\text{cos A sin A}}\end{aligned}$$

By formula,

sin² A = 1 - cos² A

$$\begin{aligned}\Rightarrow \frac{1 - \text{cos}^2 A - \text{cos}^2 A}{\text{sin A cos A}} \\[1em] \Rightarrow \frac{1 - \text{2 cos}^2 A}{\text{sin A cos A}}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that tan A - cot A = $\frac{\text{1 - 2 cos}^2 A}{\text{sin A cos A}}$

Solving L.H.S. of the equation :

⇒ cosec⁴ A - cosec² A

⇒ cosec² A(cosec² A - 1)

By formula,

cosec² A = 1 + cot² A

⇒ (1 + cot² A)(1 + cot² A - 1)

⇒ (1 + cot² A)cot² A

⇒ cot² A + cot⁴ A.

Since, L.H.S. = R.H.S.

Hence, proved that cosec⁴ A - cosec² A = cot⁴ A + cot² A.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \text{sec A(1 - sin A)(sec A + tan A)} \\[1em] \Rightarrow \frac{1}{\text{cos A}} \times (\text{1 - sin A}) \times \Big(\frac{1}{\text{cos A}} + \frac{\text{sin A}}{\text{cos A}}\Big) \\[1em] \Rightarrow \frac{1}{\text{cos A}} \times (\text{1 - sin A}) \times \Big(\frac{\text{1 + sin A}}{\text{cos A}}\Big) \\[1em] \Rightarrow \frac{1 - \text{sin}^2 A}{\text{cos}^2A}\end{aligned}$$

By formula,

cos² A = 1 - sin² A

$$\begin{aligned}\Rightarrow \frac{1 - \text{sin}^2 A}{1 - \text{sin}^2A} \\[1em] \Rightarrow 1.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that sec A(1 - sin A)(sec A + tan A) = 1.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \text{sec}^2 A + \text{cosec}^2 A \\[1em] \Rightarrow \frac{1}{\text{cos}^2 A} + \frac{1}{\text{sin}^2 A} \\[1em] \Rightarrow \frac{\text{sin}^2 A + \text{cos}^2 A}{\text{cos}^2 A. \text{sin}^2 A}\end{aligned}$$

As, sin² A + cos² A = 1

$$\begin{aligned}\Rightarrow \frac{1}{\text{cos}^2 A. \text{sin}^2 A} \\[1em] \Rightarrow \text{sec}^2 A. \text{cosec}^2 A\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that sec² A + cosec² A = sec² A . cosec² A.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\Big(1 + \frac{\text{sin}^2 A}{\text{cos}^2 A}\Big) \times \frac{\text{cos A}}{\text{sin A}}}{\frac{1}{\text{sin}^2 A}} \\[1em] \Rightarrow \Big(\frac{\text{cos}^2 A + \text{sin}^2 A}{\text{cos}^2 A}\Big) \times \frac{\text{cos A}}{\text{sin A}} \times \text{sin}^2 A\end{aligned}$$

By formula,

cos² A + sin² A = 1

$$\begin{aligned}\Rightarrow \frac{1}{\text{cos}^2 A} \times \text{cos A. sin A} \\[1em] \Rightarrow \frac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \text{tan A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{(1 + tan}^2 A)\text{cot A}}{\text{cosec}^2 A} = \text{tan A}$ .

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \text{tan}^2 A - \text{sin}^2 A \\[1em] \Rightarrow \frac{\text{sin}^2 A}{\text{cos}^2 A} - \text{sin}^2 A \\[1em] \Rightarrow \text{sin}^2 A \times \Big(\frac{1}{\text{cos}^2 A} - 1\Big) \\[1em] \Rightarrow \text{sin}^2 A \times \Big(\frac{1 - \text{cos}^2 A}{\text{cos}^2 A}\Big)\end{aligned}$$

By formula,

1 - cos²A = sin² A.

$$\begin{aligned}\Rightarrow \text{sin}^2 A \times \frac{\text{sin}^2 A}{\text{cos}^2 A} \\[1em] \Rightarrow \text{sin}^2 A. \text{tan}^2 A\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that tan² A - sin² A = tan² A. sin² A.

By formula,

cosec² A = 1 + cot² A

sin² A = 1 - cos² A

Solving L.H.S. of the equation

⇒ (cosec A + sin A)(cosec A - sin A)

⇒ cosec² A - sin² A

⇒ 1 + cot² A - (1 - cos² A)

⇒ 1 - 1 + cot² A + cos² A

⇒ cot² A + cos² A.

Hence, proved that (cosec A + sin A)(cosec A - sin A) = cot² A + cos² A.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \text{(cosec A - sin A)(sec A - cos A)(tan A + cot A)} \\[1em] \Rightarrow \Big(\frac{1}{\text{sin A}} - \text{sin A}\Big) \times \Big(\frac{1}{\text{cos A}} - \text{cos A}\Big) \times \Big(\frac{\text{sin A}}{\text{cos A}} + \frac{\text{cos A}}{\text{sin A}}\Big) \\[1em] \Rightarrow \Big(\frac{1 - \text{sin}^2 A}{\text{sin A}}\Big) \times \Big(\frac{1 - \text{cos}^2 A}{\text{cos A}}\Big) \times \Big(\frac{\text{sin}^2 A + \text{cos}^2 A}{\text{cos A. sin A}}\Big)\end{aligned}$$

By formula,

1 - sin² A = cos² A, 1 - cos² A = sin² A and sin² A + cos² A = 1.

$$\begin{aligned}\Rightarrow \Big(\frac{\text{cos}^2 A}{\text{sin A}} \times \frac{\text{sin}^2 A}{\text{cos A}} \times \frac{1}{\text{cos A. sin A}}\Big) \\[1em] \Rightarrow \frac{\text{cos}^2 A. \text{sin}^2 A}{\text{cos}^2 A. \text{sin}^2 A} \\[1em] \Rightarrow 1.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that (cosec A - sin A)(sec A - cos A)(tan A + cot A) = 1.

By formula,

sin² A + cos² A = 1

sec² A = 1 + tan² A

cosec² A = 1 + cot² A

Solving L.H.S. of the equation :

⇒ (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A

⇒ sin² A + cosec² A + 2 sin A. cosec A + cos² A + sec² A + 2 cos A. sec A

⇒ sin² A + 1 + cot² A + 2 × sin A × $\frac{1}{\text{sin A}}$ + cos² A + 1 + tan² A + 2 × cos A × $\frac{1}{\text{cos A}}$

⇒ sin² A + cos² A + 1 + cot² A + 2 + 1 + tan² A + 2

⇒ 1 + 1 + 2 + 1 + 2 + cot² A + tan² A

⇒ 7 + tan² A + cot² A.

Since, L.H.S. = R.H.S.

Hence, proved that (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A.

Solving L.H.S. of the equation :

⇒ sec² A . cosec² A

⇒ $\frac{1}{\text{cos}^2 A. \text{ sin}^2 A}$.

Solving R.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{sin}^2 A}{\text{cos}^2 A} + \frac{\text{cos}^2 A}{\text{sin}^2 A} + 2 \\[1em] \Rightarrow \frac{\text{sin}^4 A + \text{cos}^4 A + \text{2 sin}^2 A \text{ cos}^2 A}{\text{cos}^2 A \text{ sin}^2 A} \\[1em] \Rightarrow \frac{(\text{sin}^2 A + \text{cos}^2 A)^2}{\text{cos}^2 A \text{ sin}^2 A}\end{aligned}$$

By formula,

sin² A + cos² A = 1.

$$\begin{aligned}\Rightarrow \frac{(1)^2}{\text{cos}^2 A \text{ sin}^2 A} \\[1em] \Rightarrow \frac{1}{\text{cos}^2 A \text{ sin}^2 A}.\end{aligned}$$

Since, L.H.S. = R.H.S. = $\frac{1}{\text{cos}^2 A \text{ sin}^2 A}.$

Hence, proved that sec² A . cosec² A = tan² A + cot² A + 2.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{cosec A}}{\text{cosec A - 1}} + \frac{\text{cosec A}}{\text{cosec A + 1}} \\[1em] \Rightarrow \frac{\text{cosec A(cosec A + 1) + cosec A(cosec A - 1)}}{\text{(cosec A - 1)(cosec A + 1)}} \\[1em] \Rightarrow \frac{\text{cosec}^2 A + \text{cosec A + cosec}^2 A - \text{cosec A}}{\text{cosec}^2 A - 1} \\[1em] \Rightarrow \frac{\text{2 cosec}^2 A}{\text{cot}^2 A} \\[1em] \Rightarrow \frac{2 \times \frac{1}{\text{sin}^2 A}}{\frac{\text{cos}^2 A}{\text{sin}^2 A}} \\[1em] \Rightarrow \frac{2 \times \frac{1}{\text{sin}^2 A} \times \text{sin}^2 A}{\text{cos}^2 A} \\[1em] \Rightarrow \frac{2}{\text{cos}^2 A} \\[1em] \Rightarrow 2\text{sec}^2 A.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{cosec A}}{\text{cosec A - 1}} + \frac{\text{cosec A}}{\text{cosec A + 1}}$ = 2 sec² A.

Solving R.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{tan}^2 A}{\text{(sec A - 1)}^2} \\[1em] \Rightarrow \frac{\frac{\text{sin}^2 A}{\text{cos}^2 A}}{\Big(\frac{1}{\text{cos A}} - 1\Big)^2} \\[1em] \Rightarrow \frac{\frac{\text{sin}^2 A}{\text{cos}^2 A}}{\Big(\frac{\text{1 - cos A}}{\text{cos} A}\Big)^2} \\[1em] \Rightarrow \frac{\frac{\text{sin}^2 A}{\text{cos}^2 A} \times \text{cos}^2 A}{\text{(1 - cos A)}^2} \\[1em] \Rightarrow \frac{\text{sin}^2 A}{\text{(1 - cos A)}^2}.\end{aligned}$$

By formula,

sin² A = 1 - cos² A

$$\begin{aligned}\Rightarrow \frac{\text{1 - cos}^2 A}{(1 - \text{ cos A})^2} \\[1em] \Rightarrow \frac{\text{(1 - cos A)(1 + cos A)}}{\text{(1 - cos A)}^2} \\[1em] \Rightarrow \frac{\text{(1 + cos A)}}{\text{(1 - cos A)}}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{1 + cos A}}{\text{1 - cos A}} = \frac{\text{tan}^2 A}{\text{(sec A - 1)}^2}$.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{1 + sin A}}{\text{cos A}} + \frac{\text{cos A}}{\text{1 + sin A}} \\[1em] \Rightarrow \frac{\text{(1 + sin A)}^2 + \text{cos}^2 A}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \frac{\text{1 + 2 sin A + sin}^2 A + \text{cos}^2 A}{\text{cos A(1 + sin A)}}\end{aligned}$$

By formula,

sin² A + cos² A = 1.

$$\begin{aligned}\Rightarrow \frac{\text{1 + 2 sin A + 1}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \frac{\text{2 + 2 sin A}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \frac{\text{2(1 + sin A)}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \frac{2}{\text{cos A}} \\[1em] \Rightarrow 2\text{ sec A}\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{1 + sin A}}{\text{cos A}} + \frac{\text{cos A}}{\text{1 + sin A}}$ = 2 sec A.

Solving R.H.S. of the equation :

$$\begin{aligned}\Rightarrow \text{(sec A - tan A)}^2 \\[1em] \Rightarrow \Big(\frac{1}{\text{cos A}} - \frac{\text{sin A}}{\text{cos A}}\Big)^2 \\[1em] \Rightarrow \Big(\frac{\text{1 - sin A}}{\text{cos A}}\Big)^2 \\[1em] \Rightarrow \frac{\text{(1 - sin A)}^2}{\text{cos}^2 A}\end{aligned}$$

By formula,

cos² A = 1 - sin² A

$$\begin{aligned}\Rightarrow \frac{\text{(1 - sin A)}^2}{\text{1 - sin}^2 A} \\[1em] \Rightarrow \frac{\text{(1 - sin A)}^2}{\text{(1 - sin A)(1 + sin A)}} \\[1em] \Rightarrow \frac{\text{1 - sin A}}{\text{1 + sin A}}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{1 - sin A}}{\text{1 + sin A}}$ = (sec A - tan A)².

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\frac{1}{\text{sin A}} - 1}{\frac{1}{\text{sin A}} + 1} \\[1em] \Rightarrow \frac{\frac{\text{1 - sin A}}{\text{sin A}}}{\frac{\text{1 + sin A}}{\text{sin A}}} \\[1em] \Rightarrow \frac{\text{(1 - sin A)} \times \text{ sin A}}{\text{(1 + sin A)} \times \text{ sin A}} \\[1em] \Rightarrow \frac{\text{1 - sin A}}{\text{1 + sin A}}.\end{aligned}$$

Solving R.H.S. of the equation :

$$\begin{aligned}\Rightarrow \Big(\frac{\text{cos A}}{\text{1 + sin A}}\Big)^2 \\[1em] \Rightarrow \frac{\text{cos}^2 A}{\text{(1 + sin A)}^2} \\[1em] \Rightarrow \frac{\text{1 - sin}^2 A}{\text{(1 + sin A)}^2} \\[1em] \Rightarrow \frac{\text{(1 - sin A)(1 + sin A)}}{\text{(1 + sin A)}^2} \\[1em] \Rightarrow \frac{\text{(1 - sin A)}}{\text{(1 + sin A)}}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{cosec A - 1}}{\text{cosec A + 1}} = \Big(\frac{\text{cos A}}{\text{1 + sin A}}\Big)^2$.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{sin}^2 A}{\text{cos}^2 A} - \frac{\text{sin}^2 B}{\text{cos}^2 B} \\[1em] \Rightarrow \frac{\text{sin}^2 A. \text{ cos}^2 B - \text{sin}^2 B. \text{ cos}^2 A}{\text{cos}^2 A. \text{ cos}^2 B} \\[1em] \Rightarrow \frac{\text{sin}^2 A(1 - \text{ sin}^2 B) - \text{sin}^2 B(1 - \text{sin}^2 A)}{\text{cos}^2 A. \text{ cos}^2 B} \\[1em] \Rightarrow \frac{\text{sin}^2 A - \text{sin}^2 A. \text{ sin}^2 B - \text{sin}^2 B + \text{ sin}^2 A. \text{ sin}^2 B}{\text{cos}^2 A. \text{ cos}^2 B} \\[1em] \Rightarrow \frac{\text{sin}^2 A - \text{ sin}^2 B}{\text{cos}^2 A. \text{ cos}^2 B}\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that tan² A - tan² B = $\frac{\text{sin}^2 A - \text{sin}^2 B}{\text{cos}^2 A. \text{cos}^2 B}$.

Solving L.H.S. of the equation :

$\Rightarrow \frac{\text{sin θ(1 - 2 sin}^2 θ)}{\text{cos θ(2 cos}^2 θ - 1)}$

By formula,

sin² θ = 1 - cos² θ

$$\begin{aligned}\Rightarrow \frac{\text{sin θ[1 - 2(1 - cos}^2 θ)]}{\text{cos θ(2 cos}^2 θ - 1)} \\[1em] \Rightarrow \text{tan θ} \times \frac{1 - 2 + \text{ 2 cos}^2 θ}{\text{(2 cos}^2 θ - 1)} \\[1em] \Rightarrow \text{tan θ} \times \frac{\text{2 cos}^2 θ - 1}{\text{2 cos}^2 θ - 1} \\[1em] \Rightarrow \text{tan θ}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{sin θ - 2 sin}^3 θ}{\text{2 cos}^3 θ - \text{ cos θ}}$ = tan θ.

Solving L.H.S. of the equation :

$\Rightarrow \frac{\text{cos A}}{\text{1 - sin A}}$

Multiplying numerator and denominator by (1 + sin A)

$$\begin{aligned}\Rightarrow \frac{\text{cos A(1 + sin A)}}{\text{(1 - sin A)(1 + sin A)}} \\[1em] \Rightarrow \frac{\text{cos A(1 + sin A)}}{\text{1 - sin}^2 A}\end{aligned}$$

By formula,

cos² A = 1 - sin² A

$$\begin{aligned}\Rightarrow \frac{\text{cos A(1 + sin A)}}{\text{cos}^2 A} \\[1em] \Rightarrow \frac{\text{1 + sin A}}{\text{\text{cos A}}} \\[1em] \Rightarrow \frac{1}{\text{cos A}} + \frac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \text{sec A + tan A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{cos A}}{\text{1 - sin A}}$ = sec A + tan A.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{sin A} \times \frac{\text{sin A}}{\text{cos A}}}{\text{1 - cos A}} \\[1em] \Rightarrow \frac{\text{sin}^2 A}{\text{cos A(1 - cos A)}}\end{aligned}$$

By formula,

sin² A = 1 - cos² A

$$\begin{aligned}\Rightarrow \frac{1 - \text{cos}^2 A}{\text{cos A(1 - cos A)}} \\[1em] \Rightarrow \frac{\text{(1 - cos A)(1 + cos A)}}{\text{cos A(1 - cos A)}} \\[1em] \Rightarrow \frac{\text{1 + cos A}}{\text{cos A}} \\[1em] \Rightarrow \frac{1}{\text{cos A}} + \frac{\text{cos A}}{\text{cos A}} \\[1em] \Rightarrow \text{sec A + 1}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{sin A tan A}}{\text{1 - cos A}}$ = 1 + sec A.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \Big(1 + \frac{\text{cos A}}{\text{sin A}} - \frac{1}{\text{sin A}}\Big)\Big(1 + \frac{\text{sin A}}{\text{cos A}}+ \frac{1}{\text{cos A}}\Big) \\[1em] \Rightarrow \Big(\frac{\text{sin A + cos A - 1}}{\text{sin A}}\Big)\Big(\frac{\text{cos A + sin A + 1}}{\text{cos A}}\Big) \\[1em] \Rightarrow \frac{\text{(sin A + cos A - 1)(sin A + cos A + 1)}}{\text{sin A cos A}} \\[1em] \Rightarrow \frac{\text{sin}^2 A + \text{sin A cos A + sin A + cos A sin A + cos A} + \text{ cos}^2 A - \text{sin A - cos A - 1}}{\text{sin A cos A}} \\[1em] \Rightarrow \frac{\text{sin}^2 A + \text{cos}^2 A + \text{2 cos A sin A - 1}}{\text{sin A cos A}}.\end{aligned}$$

By formula,

sin² A + cos² A = 1.

$$\begin{aligned}\Rightarrow \frac{1 + \text{2 cos A sin A - 1}}{\text{sin A cos A}} \\[1em] \Rightarrow \frac{\text{2 cos A sin A}}{\text{cos A sin A}} \\[1em] \Rightarrow 2.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that (1 + cot A - cosec A)(1 + tan A + sec A) = 2.

Solving L.H.S. of the equation :

$\Rightarrow \sqrt{\frac{\text{1 + sin A}}{\text{1 - sin A}}}$

Multiplying numerator and denominator by $\sqrt{1 + \text{sin A}}$

$$\begin{aligned}\Rightarrow \sqrt{\frac{\text{1 + sin A}}{\text{1 - sin A}}} \times \sqrt{\frac{\text{1 + sin A}}{\text{1 + sin A}}} \\[1em] \Rightarrow \sqrt{\frac{\text{(1 + sin A)(1 + sin A)}}{\text{(1 - sin A)(1 + sin A)}}} \\[1em] \Rightarrow \sqrt{\frac{\text{(1 + sin A)(1 + sin A)}}{\text{1 - sin}^2 \text{A}}} \\[1em] \Rightarrow \sqrt{\frac{\text{(1 + sin A)}^2}{\text{cos}^2 A}} \\[1em] \Rightarrow \frac{\text{1 + sin A}}{\text{cos A}} \\[1em] \Rightarrow \frac{1}{\text{cos A}} + \frac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \text{sec A + tan A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\sqrt{\frac{\text{1 + sin A}}{\text{1 - sin A}}}$ = sec A + tan A.

Solving L.H.S. of the equation :

$\Rightarrow \sqrt{\frac{\text{1 - cos A}}{\text{1 + cos A}}}$ = cosec A - cot A

Multiplying numerator and denominator by $\sqrt{1 - \text{cos A}}$

$$\begin{aligned}\Rightarrow \sqrt{\frac{\text{1 - cos A}}{\text{1 + cos A}}} \times \sqrt{\frac{\text{1 - cos A}}{\text{1 - cos A}}} \\[1em] \Rightarrow \sqrt{\frac{\text{(1 - cos A)(1 - cos A)}}{\text{(1 + cos A)(1 - cos A)}}} \\[1em] \Rightarrow \sqrt{\frac{\text{(1 - cos A)}^2}{\text{(1 - cos}^2 A)}}\end{aligned}$$

By formula,

1 - cos² A = sin² A

$$\begin{aligned}\Rightarrow \sqrt{\frac{\text{(1 - cos A)}^2}{\text{sin}^2 A}} \\[1em] \Rightarrow \frac{\text{1 - cos A}}{\text{sin A}} \\[1em] \Rightarrow \frac{1}{\text{sin A}} - \frac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \text{cosec A - cot A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\sqrt{\frac{\text{1 - cos A}}{\text{1 + cos A}}}$ = cosec A - cot A.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow 1 - \frac{\text{cos}^2 A}{\text{1 + sin A}} \\[1em] \Rightarrow \frac{\text{1 + sin A - cos}^2 A}{\text{1 + sin A}}\end{aligned}$$

By formula,

cos² A = 1 - sin² A

$$\begin{aligned}\Rightarrow \frac{\text{1 + sin A - (1 - sin}^2 A)}{\text{1 + sin A}} \\[1em] \Rightarrow \frac{\text{1 + sin A - 1 + sin}^2 A}{\text{1 + sin A}} \\[1em] \Rightarrow \frac{\text{sin A + sin}^2 A}{\text{1 + sin A}} \\[1em] \Rightarrow \frac{\text{sin A(1 + sin A)}}{\text{1 + sin A}} \\[1em] \Rightarrow \text{sin A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that 1 - $\frac{\text{cos}^2 A}{\text{1 + sin A}}$ = sin A.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{sin A - cos A + sin A + cos A}}{\text{(sin A + cos A)(sin A - cos A)}} \\[1em] \Rightarrow \frac{\text{2 sin A}}{\text{sin}^2 A - \text{cos}^2 A}\end{aligned}$$

By formula,

sin² A = 1 - cos² A

$$\begin{aligned}\Rightarrow \frac{\text{2 sin A}}{\text{1 - cos}^2 A - \text{cos}^2 A} \\[1em] \Rightarrow \frac{\text{2 sin A}}{\text{1 - 2 cos}^2 A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{1}{\text{sin A + cos A}} + \frac{1}{\text{sin A - cos A}} = \frac{\text{2 sin A}}{\text{1 - 2 cos}^2 A}$.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{(sin A + cos A)}^2 + \text{(sin A - cos A)}^2}{\text{(sin A - cos A)(sin A + cos A)}} \\[1em] \Rightarrow \frac{\text{sin}^2 A + \text{cos}^2 + \text{2 sin A cos A} + \text{sin}^2 A + \text{cos}^2 A - \text{2 sin A cos A}}{\text{sin}^2 A - \text{cos}^2 A} \\[1em] \Rightarrow \frac{\text{2 (sin}^2 A + \text{ cos}^2 A)}{\text{sin}^2 A - \text{cos}^2 A}\end{aligned}$$

By formula,

sin² A + cos² A = 1

cos² A = 1 - sin² A

$$\begin{aligned}\Rightarrow \frac{2}{\text{sin}^2 A - (1 - \text{sin}^2 A)} \\[1em] \Rightarrow \frac{2}{\text{sin}^2 A - 1 + \text{sin}^2 A} \\[1em] \Rightarrow \frac{2}{\text{2 sin}^2 A - 1}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{sin A + \text{cos A}}}{\text{sin A - cos A}} + \frac{\text{sin A - cos A}}{\text{sin A + cos A}} = \frac{2}{\text{2 sin}^2 A - 1}$

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{1 + sin A}}{\text{cosec A - cot A}} - \frac{\text{1 - sin A}}{\text{cosec A + cot A}} \\[1em] \Rightarrow \frac{\text{(1 + sin A)(cosec A + cot A)} - \text{(1 - sin A)(cosec A - cot A)}}{\text{cosec}^2 A - \text{cot}^2 A}\end{aligned}$$

By formula,

cosec² A - cot² A = 1

⇒ (1 + sin A)(cosec A + cot A) - (1 - sin A)(cosec A - cot A)

⇒ cosec A + cot A + sin A cosec A + sin A cot A - (cosec A - cot A - sin A cosec A + sin A cot A)

⇒ cosec A - cosec A + cot A + cot A + sin A cosec A + sin A cosec A + sin A cot A - sin A cot A

⇒ 2 cot A + 2 sin A cosec A

⇒ 2 cot A + $2 \text{ sin A} \times \frac{1}{\text{sin A}}$

⇒ 2 cot A + 2

⇒ 2(cot A + 1).

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{1 + sin A}}{\text{cosec A - cot A}} - \frac{\text{1 - sin A}}{\text{cosec A + cot A}}$ = 2(1 + cot A).

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{cos θ} \times \frac{\text{cos θ}}{\text{sin θ}}}{\text{(1 + sin θ)}} \\[1em] \Rightarrow \frac{\text{cos}^2 θ}{\text{sin θ(1 + sin θ)}}\end{aligned}$$

By formula,

cos² θ = 1 - sin² θ

$$\begin{aligned}\Rightarrow \frac{\text{1 - sin}^2 θ}{\text{sin θ(1 + sin θ)}} \\[1em] \Rightarrow \frac{\text{(1 - sin θ)(1 + sin θ)}}{\text{sin θ(1 + sin θ)}} \\[1em] \Rightarrow \frac{\text{1 - sin θ}}{\text{sin θ}} \\[1em] \Rightarrow \frac{1}{\text{sin θ}} - \frac{\text{sin θ}}{\text{sin θ}} \\[1em] \Rightarrow \text{cosec θ - 1}\end{aligned}$$

Since, L.H.S. = R.H.S

Hence, proved that $\frac{\text{cos θ cot θ}}{\text{1 + sin θ}}$ = cosec θ - 1.

Solving,

⇒ (1 + cot² A) + (1 + tan² A)

⇒ cosec² A + sec² A

⇒ $\frac{1}{\text{sin}^2 A} + \frac{1}{\text{cos}^2 A}$

⇒ $\frac{\text{cos}^2 A + \text{sin}^2 A}{\text{sin}^2 A. \text{cos}^2 A}$

Substituting, sin² A + cos² A = 1, we get :

⇒ $\frac{1}{\text{sin}^2 A. \text{cos}^2 A}$

⇒ cosec² A. sec² A

Hence, Option 4 is the correct option.

Substituting value of a and b in L.H.S. of option 3, we get :

⇒ b² - a²

⇒ sec² A - tan² A

⇒ 1.

Since, L.H.S. = R.H.S.

Hence, Option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{1}{\text{cot θ + tan θ}} \\[1em] \Rightarrow \frac{1}{\frac{\text{cos θ}}{\text{sin θ}} + \frac{\text{sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \frac{1}{\frac{\text{cos}^2 θ + \text{sin}^2 θ}{\text{cos θ. sin θ}}} \\[1em] \Rightarrow \frac{\text{cos θ. sin θ}}{\text{cos}^2 θ + \text{sin}^2 θ}\end{aligned}$$

Substituting, sin² θ + cos² θ = 1, we get :

⇒ sin θ. cos θ

Hence, Option 2 is the correct option.

Solving,

⇒ (sec θ - cos θ)² - (sec θ + cos θ)²

⇒ (sec θ - cos θ + sec θ + cos θ)[(sec θ - cos θ) - (sec θ + cos θ)] [∵ a² - b² = (a + b)(a - b)]

⇒ (sec θ - cos θ + sec θ + cos θ)(sec θ - sec θ - cos θ - cos θ)

⇒ 2 sec θ.(-2 cos θ)

⇒ -4.sec θ.cos θ

⇒ $-4 \times \frac{1}{\text{cos θ}} \times \text{cos θ}$

⇒ -4.

Hence, Option 4 is the correct option.

Solving,

⇒ (cot A - cot B)² + (1 + cot A cot B)²

⇒ cot² A + cot² B - 2 cot A cot B + 1 + cot² A . cot² B + 2 cot A cot B

⇒ cot² A + cot² A . cot² B + 1 + cot² B

⇒ cot² A(1 + cot² B) + (1 + cot² B)

⇒ (1 + cot² B)(1 + cot² A)

⇒ $\Big(1 + \frac{\text{cos}^2 B}{\text{sin}^2 B}\Big)\Big(1 + \frac{\text{cos}^2 A}{\text{sin}^2 A}\Big)$

⇒ $\Big(\frac{\text{sin}^2 B + \text{cos}^2 B}{\text{sin}^2 B}\Big)\Big(\frac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin}^2 A}\Big)$

Substituting, sin² θ + cos² θ = 1, we get :

⇒ $\Big(\frac{1}{{\text{sin}^2 B}}\Big)\Big(\frac{1}{\text{sin}^2 A}\Big)$

⇒ cosec² B . cosec² A

Hence, Option 4 is the correct option.

To prove:

(sec A - tan A)² (1 + sin A) = (1 - sin A)

Solving L.H.S. of the above equation,

$$\begin{aligned}\Rightarrow \text{(sec A - tan A)}^2 \text{(1 + sin A)} \\[1em] \Rightarrow \Big(\frac{1}{\text{cos A}} - \frac{\text{sin A}}{\text{cos A}}\Big)^2 \text{(1 + sin A)} \\[1em] \Rightarrow \Big(\frac{\text{1 - sin A}}{\text{cos A}}\Big)^2\text{(1 + sin A)} \\[1em] \Rightarrow \frac{(1 - \text{sin A})^2\text{(1 + sin A)}}{\text{cos}^2 A} \\[1em] \Rightarrow \frac{(1 - \text{sin A})^2\text{(1 + sin A)}}{1 - \text{sin}^2 A} \\[1em] \Rightarrow \frac{(1 - \text{sin A})^2\text{(1 + sin A)}}{\text{(1 - sin A)(1 + sin A)}} \\[1em] \Rightarrow \text{1 - sin A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that (sec A - tan A)² (1 + sin A) = (1 - sin A).

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{(\text{cos}^3 A + \text{sin}^3 A)(\text{cos A - sin A}) + (\text{cos}^3 A - \text{sin}^3 A)(\text{cos A + sin A})}{\text{(cos A + sin A)(cos A - sin A)}} \\[1em] \Rightarrow \frac{\text{cos}^4 A - \text{cos}^3 A\text{ sin A} + \text{cos A sin}^3 A - \text{ sin}^4 A + \text{cos}^4 A + \text{cos}^3 A\text{ sin A} - \text{ sin}^3 A\text{ cos A} - \text{sin}^4 A}{\text{cos}^2 A - \text{sin}^2 A} \\[1em] \Rightarrow \frac{\text{2 cos}^4 A - \text{2 sin}^4 A}{\text{cos}^2 A - \text{sin}^2 A} \\[1em] \Rightarrow \frac{\text{2(cos}^4 A - \text{ sin}^4 A)}{\text{cos}^2 A - \text{sin}^2 A} \\[1em] \Rightarrow \frac{\text{2(cos}^2 A - \text{sin}^2 A)\text{(cos}^2 A + \text{sin}^2 A)}{\text{cos}^2 A - \text{sin}^2 A} \\[1em] \Rightarrow \text{2 (cos}^2 A + \text{sin}^2 A).\end{aligned}$$

By formula,

cos² A + sin² A = 1

⇒ 2 × 1 = 2.

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{cos}^3 A + \text{sin}^3 A}{\text{cos A+ sin A}} + \frac{\text{cos}^3 A - \text{sin}^3 A}{\text{cos A - sin A}}$ = 2.

By solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{tan A}}{\text{1 - cot A}} + \frac{\text{cot A}}{\text{1 - tan A}} \\[1em] \Rightarrow \frac{\text{tan A}}{1 - \frac{1}{\text{tan A}}} + \frac{\frac{1}{\text{tan A}}}{\text{1 - tan A}} \\[1em] \Rightarrow \frac{\text{tan A}}{\frac{\text{tan A - 1}}{\text{tan A}}} + \frac{1}{\text{tan A(1 - tan A)}} \\[1em] \Rightarrow \frac{\text{tan}^2 A}{\text{tan A - 1}} + \frac{1}{\text{tan A(1 - tan A)}} \\[1em] \Rightarrow \frac{\text{tan}^2 A}{\text{tan A - 1}} - \frac{1}{\text{tan A(tan A - 1)}} \\[1em] \Rightarrow \frac{\text{tan}^3 A - 1}{\text{tan A(tan A - 1)}} \\[1em] \Rightarrow \frac{\text{(tan A - 1)(tan}^2 A + \text{ tan A + 1)}}{\text{tan A(tan A - 1)}} \\[1em] \Rightarrow \frac{\text{tan}^2 A + \text{ tan A + 1}}{\text{tan A}} \\[1em] \Rightarrow \frac{\text{tan}^2 A}{\text{tan A}} + \frac{\text{tan A}}{\text{tan A}} + \frac{1}{\text{tan A}} \\[1em] \Rightarrow \text{tan A + 1 + cot A} \\[1em] \Rightarrow \frac{\text{sin A}}{\text{cos A}} + 1 + \frac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \frac{\text{sin}^2 A + \text{sin A cos A + cos}^2 A}{\text{sin A cos A}}.\end{aligned}$$

By formula,

sin² A + cos² A = 1

$$\begin{aligned}\Rightarrow \frac{1 + \text{sin A cos A}}{\text{sin A cos A}} \\[1em] \Rightarrow \Rightarrow \frac{1}{\text{sin A cos A}} + \frac{\text{sin A cos A}}{\text{sin A cos A}} \\[1em] \Rightarrow \text{cosec A sec A} + 1.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{tan A}}{\text{1 - cot A}} + \frac{\text{cot A}}{\text{1 - tan A}}$ = sec A cosec A + 1.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \Big(\text{tan A} + \frac{1}{\text{cos A}}\Big)^2 + \Big(\text{tan A} - \frac{1}{\text{cos A}}\Big)^2 \\[1em] \Rightarrow \Big(\frac{\text{sin A}}{\text{cos A}} + \frac{1}{\text{cos A}}\Big)^2 + \Big(\frac{\text{sin A}}{\text{cos A}} - \frac{1}{\text{cos A}}\Big)^2 \\[1em] \Rightarrow \Big(\frac{\text{sin A + 1}}{\text{cos A}}\Big)^2 + \Big(\frac{\text{sin A - 1}}{\text{cos A}}\Big)^2 \\[1em] \Rightarrow \frac{\text{sin}^2 A + 1 + \text{2 sin A}}{\text{cos}^2 A} + \frac{\text{sin}^2 A + 1 - \text{2 sin A}}{\text{cos}^2 A} \\[1em] \Rightarrow \frac{\text{sin}^2 A + 1 + \text{2 sin A} + \text{sin}^2 A + 1 - \text{2 sin A}}{\text{cos}^2 A} \\[1em] \Rightarrow \frac{\text{2(1 + sin}^2 A)}{\text{cos}^2 A}\end{aligned}$$

By formula,

cos² A = 1 - sin² A

$\Rightarrow 2\Big(\frac{1 + \text{sin}^2 A}{1 - \text{sin}^2 A}\Big)$

Since, L.H.S. = R.H.S.

Hence, proved that

$\Big(\text{tan A} + \frac{1}{\text{cos A}}\Big)^2 + \Big(\text{tan A} - \frac{1}{\text{cos A}}\Big)^2 = 2\Big(\frac{\text{1 + sin}^2 A}{\text{1 - sin}^2 A}\Big)$.

Solving L.H.S. of the equation :

⇒ 2 sin² A + cos⁴ A

⇒ 2 sin² A + (cos² A)²

By formula,

cos² A = 1 - sin² A

⇒ 2 sin² A + (1 - sin² A)²

⇒ 2 sin² A + 1 + sin⁴ A - 2 sin² A

⇒ 1 + sin⁴ A.

Since, L.H.S. = R.H.S.

Hence, proved that 2 sin² A + cos⁴ A = 1 + sin⁴ A.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{(sin A - sin B)(sin A + sin B) + (cos A - cos B)(cos A + cos B)}}{\text{(cos A + cos B)(sin A + sin B)}} = 0 \\[1em] \Rightarrow \frac{\text{sin}^2 A - \text{sin}^2 B + \text{cos}^2 A - \text{cos}^2 B}{\text{(cos A + cos B)(sin A + sin B)}} \\[1em] \Rightarrow \frac{\text{sin}^2 A + \text{cos}^2 A -\text{(sin}^2 B + \text{cos}^2 B)}{\text{(cos A + cos B)(sin A + sin B)}}\end{aligned}$$

By formula,

sin² θ + cos² θ = 1

$$\begin{aligned}\therefore \frac{1 - 1}{\text{(cos A + cos B)(sin A + sin B)}} \\[1em] \Rightarrow 0.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{sin A - sin B}}{\text{cos A + cos B}} + \frac{\text{cos A - cos B}}{\text{sin A + sin B}}$ = 0.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \text{(cosec A - sin A)(sec A - cos A)} \\[1em] \Rightarrow \Big(\frac{1}{\text{sin A}} - \text{sin A}\Big) \times \Big(\frac{1}{\text{cos A}} - \text{cos A}\Big) \\[1em] \Rightarrow \Big(\frac{\text{1 - sin}^2 A}{\text{sin A}}\Big) \times \Big(\frac{\text{1 - cos}^2 A}{\text{cos A}}\Big)\end{aligned}$$

By formula,

1 - sin² A = cos² A

1 - cos² A = sin² A

$$\begin{aligned}\Rightarrow \frac{\text{cos}^2 A}{\text{sin A}} \times \frac{\text{sin}^2 A}{\text{cos A}} \\[1em] \Rightarrow \text{cos A sin A}.\end{aligned}$$

Solving R.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{1}{\text{tan A + cot A}} \\[1em] \Rightarrow \frac{1}{\frac{\text{sin A}}{\text{cos A}} + \frac{\text{cos A}}{\text{sin A}}} \\[1em] \Rightarrow \frac{1}{\frac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}}} \\[1em] \Rightarrow \frac{\text{sin A cos A}}{\text{sin}^2 A + \text{cos}^2 A}\end{aligned}$$

By formula,

sin² θ + cos² θ = 1

$$\begin{aligned}\Rightarrow \frac{\text{sin A cos A}}{1} \\[1em] \Rightarrow \text{sin A cos A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that (cosec A - sin A)(sec A - cos A) = $\frac{1}{\text{tan A + cot A}}$.

By formula,

sec² θ = 1 + tan² θ

Solving L.H.S. of the equation :

⇒ (1 + tan A. tan B)² + (tan A - tan B)²

⇒ 1 + tan² A tan² B + 2 tan A tan B + tan² A + tan² B - 2 tan A tan B

⇒ 1 + tan² A tan² B + tan² A + tan² B

⇒ 1 + tan² A + tan² A tan² B + tan² B

⇒ sec² A + tan² B(tan² A + 1)

⇒ sec² A + tan² B sec² A

⇒ sec² A(1 + tan² B)

⇒ sec² A sec² B.

Since, L.H.S. = R.H.S.

Hence, proved that (1 + tan A. tan B)² + (tan A - tan B)² = sec² A sec² B.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{cos A + sin A + 1 + cos A + sin A - 1}}{\text{(cos A + sin A - 1)(cos A + sin A + 1)}} \\[1em] \Rightarrow \frac{\text{2(cos A + sin A)}}{\text{(cos A + sin A)}^2 - 1^2} \\[1em] \Rightarrow \frac{\text{2(cos A + sin A)}}{\text{cos}^2 A + \text{sin}^2 A + \text{2 cos A sin A} - 1} \\[1em] \Rightarrow \frac{\text{2(cos A + sin A)}}{1 - 1 + \text{2 cos A sin A}} \\[1em] \Rightarrow \frac{\text{2(cos A + sin A)}}{\text{2 cos A sin A}} \\[1em] \Rightarrow \frac{\text{(cos A + sin A)}}{\text{cos A sin A}} \\[1em] \Rightarrow \frac{\text{cos A}}{\text{cos A sin A}} + \frac{\text{sin A}}{\text{cos A sin A}} \\[1em] \Rightarrow \frac{1}{\text{sin A}} + \frac{1}{\text{cos A}}\\[1em] \Rightarrow \text{cosec A + sec A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{1}{\text{cos A + sin A - 1}} + \frac{1}{\text{cos A + sin A + 1}}$ = cosec A + sec A.

To prove:

x² + y² = m² + n²

Substituting value of m and n in R.H.S. of the equation :

= (x cos A + y sin A)² + (x sin A - y cos A)²

= x² cos² A + y² sin² A + 2xy cos A sin A + x² sin² A + y² cos² A - 2xy sin A cos A

= x² cos² A + x² sin² A + y² cos² A + y² sin² A

= x²(sin² A + cos² A) + y²(sin² A + cos² A)

By formula,

sin² A + cos² A = 1

⇒ x² × 1 + y² × 1

⇒ x² + y².

Since, L.H.S. = R.H.S.

Hence, proved that x² + y² = m² + n².

To prove:

m² - n² = a² - b²

Substituting value of m and n in L.H.S. of the above equation :

= (a sec A + b tan A)² - (a tan A + b sec A)²

= a² sec² A + b² tan² A + 2ab sec A tan A - (a² tan² A + b² sec² A + 2ab sec A tan A)

= a² sec² A - a² tan² A + b²tan² A - b² sec² A + 2ab sec A tan A - 2ab sec A tan A

= a² (sec² A - tan² A) + b² (tan² A - sec² A)

= a² (sec² A - tan² A) - b² (sec² A - tan² A)

By formula,

sec² A - tan² A = 1

⇒ a² × 1 - b² × 1

⇒ a² - b²

Since, L.H.S. = R.H.S.

Hence, proved that m² - n² = a² - b².

To prove:

⇒ x² + y² + z² = r²

Substituting value of x, y and z in L.H.S. of the equation :

= (r cos A cos B)² + (r cos A sin B)² + (r sin A)²

= r² cos² A cos² B + r² cos² A sin² B + r² sin² A

= r²cos² A(cos² B + sin² B) + r² sin² A

⇒ r²cos² A + r²sin² A [∵ sin² θ + cos² θ = 1]

⇒ r²(cos² A + sin² A)

⇒ r² × 1 [∵ sin² θ + cos² θ = 1]

⇒ r².

Since, L.H.S. = R.H.S.

Hence, proved that x² + y² + z² = r².

To prove:

(m² + n²) cos² B = n²

Substituting value of m and n in L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \Big[\Big(\frac{\text{cos A}}{\text{cos B}}\Big)^2 + \Big(\frac{\text{cos A}}{\text{sin B}}\Big)^2\Big]\text{cos}^2 B \\[1em] \Rightarrow \Big[\frac{\text{cos}^2 A}{\text{cos}^2 B} + \frac{\text{cos}^2 A}{\text{sin}^2 B}\Big]\text{cos}^2 B \\[1em] \Rightarrow \frac{\text{cos}^2 A \text{ sin}^2 B + \text{cos}^2 A \text{cos}^2 B}{\text{ cos}^2 B \text{ sin}^2 B} \times \text{cos}^2 B \\[1em] \Rightarrow \frac{\text{cos}^2 A(\text{sin}^2 B + \text{cos}^2 B)}{\text{sin}^2 B}\end{aligned}$$

By formula,

⇒ sin² B + cos² B = 1

$$\begin{aligned}\Rightarrow \frac{\text{cos}^2 A}{\text{sin}^2 B} \\[1em] \Rightarrow \Big(\frac{\text{cos A}}{\text{sin B}}\Big)^2 \\[1em] \Rightarrow n^2.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that (m² + n²) cos² B = n².

sin (90° - A) = cos A

To prove:

sin² A + sin² (90° - A)

⇒ sin² A + cos² A

⇒ 1.

Hence, Option 2 is the correct option.

In triangle ABC,

By angle sum property of triangle,

⇒ A + B + C = 180°

⇒ A + C = 180° - B

⇒ $\frac{A + C}{2} = \frac{180° - B}{2}$ ........(1)

Substituting value of $\frac{A + C}{2}$ in sec $\frac{C + A}{2}$, we get :

$$\begin{aligned}\Rightarrow \text{sec } \frac{C + A}{2} \\[1em] \Rightarrow \text{sec } \frac{180° - B}{2} \\[1em] \Rightarrow \text{sec } \Big(90° - \frac{B}{2}\Big) \\[1em] \Rightarrow \text{cosec } \frac{B}{2}.\end{aligned}$$

Hence, Option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{\text{cot 46°}}{\text{tan 44°}} - 3 \frac{\text{sec 20°}}{\text{cosec 70°}} + 5 \\[1em] \Rightarrow \frac{\text{cot (90° - 44°)}}{\text{tan 44°}} - 3 \frac{\text{sec (90° - 70°)}}{\text{cosec 70°}} + 5 \\[1em] \Rightarrow \frac{\text{tan 44°}}{\text{tan 44°}} - 3 \frac{\text{cosec 70°}}{\text{cosec 70°}} + 5 \\[1em] \Rightarrow 1 - 3 \times 1 + 5 \\[1em] \Rightarrow 1 - 3 + 5 \\[1em] \Rightarrow 3.\end{aligned}$$

Hence, Option 3 is the correct option.

Solving,

⇒ sin 67° . cos 23° + cos 67° . sin 23°

⇒ sin 67° . cos (90° - 67°) + cos 67° . sin (90° - 67°)

⇒ sin 67° . sin 67° + cos 67° . cos 67°

⇒ sin² 67° + cos² 67°

⇒ 1.

Hence, Option 4 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{\text{cos θ. cos (90° - θ)}}{\text{tan (90° - θ)}} \\[1em] \Rightarrow \frac{\text{cos θ . sin θ}}{\text{cot θ}} \\[1em] \Rightarrow \frac{\text{cos θ . sin θ}}{\frac{\text{cos θ}}{\text{sin θ}}} \\[1em] \Rightarrow \frac{\text{cos θ . sin θ . sin θ}}{\text{cos θ}} \\[1em] \Rightarrow \text{sin}^2 θ.\end{aligned}$$

Hence, Option 2 is the correct option.

Solving L.H.S. of the equation :

⇒ tan 10° tan 15° tan 75° tan 80°

⇒ tan 10° tan 15° tan (90 - 15)° tan (90 - 10)°

By formula,

tan (90° - A) = cot A

⇒ tan 10° tan 15° cot 15° cot 10°

⇒ tan 10° × tan 15° $\times \frac{1}{\text{tan 15°}} \times \frac{1}{\text{tan 10°}}$

⇒ 1.

Since, L.H.S. = R.H.S.

Hence, proved that tan 10° tan 15° tan 75° tan 80° = 1.

Solving L.H.S. of the equation :

⇒ sin 42° sec (90 - 42)° + cos 42° cosec (90 - 42)°

By formula,

sec (90° - A) = cosec A and cosec (90° - A) = sec A.

⇒ sin 42° cosec 42° + cos 42° sec 42°

⇒ sin 42° $\times \frac{1}{\text{sin 42°}}$ + cos 42° $\times \frac{1}{\text{cos 42°}}$

⇒ 1 + 1

⇒ 2.

Since, L.H.S. = R.H.S.

Hence, proved that sin 42° sec 48° + cos 42° cosec 48° = 2.

(i) Solving,

⇒ sin 59° + tan 63°

⇒ sin (90 - 31)° + tan (90 - 27)°

By formula,

sin (90° - A) = cos A and tan (90° - A) = cot A.

⇒ cos 31° + cot 27°.

Hence, sin 59° + tan 63° = cos 31° + cot 27°.

(ii) Solving,

⇒ cosec 68° + cot 72°

⇒ cosec (90 - 22)° + cot (90 - 18)°

By formula,

cosec (90° - A) = sec A and cot (90° - A) = tan A.

⇒ sec 22° + tan 18°.

Hence, cosec 68° + cot 72° = sec 22° + tan 18°.

To prove:

$\frac{\text{sin A}}{\text{sin(90° - A)}} + \frac{\text{cos A}}{\text{cos(90° - A)}}$ = sec A cosec A

By formula,

sin (90° - A) = cos A and cos (90° - A) = sin A.

Substituting above values in L.H.S. :

$$\begin{aligned}\Rightarrow \frac{\text{sin A}}{\text{cos A}} + \frac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \frac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}}\end{aligned}$$

By formula,

sin² A + cos² A = 1

$$\begin{aligned}\Rightarrow \frac{1}{\text{sin A cos A}} \\[1em] \Rightarrow \frac{1}{\text{sin A}} \times \frac{1}{\text{cos A}} \\[1em] \Rightarrow \text{cosec A sec A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{sin A}}{\text{sin(90° - A)}} + \frac{\text{cos A}}{\text{cos(90° - A)}}$ = sec A cosec A.

By formula,

cos (90° - A) = sin A, sec (90° - A) = cosec A, cosec (90° - A) = sec A and sin (90° - A) = cos A.

$$\begin{aligned}\Rightarrow \text{sin A cos A} - \frac{\text{sin A sin A cos A}}{\text{cosec A}} - \frac{\text{cos A cos A sin A}}{\text{sec A}} \\[1em] \Rightarrow \text{sin A cos A} - \frac{\text{sin}^2 A \text{cos A}}{\frac{1}{\text{sin A}}} - \frac{\text{cos}^2 A \text{sin A}}{\frac{1}{\text{cos A}}} \\[1em] \Rightarrow \text{sin A cos A} - \text{sin}^3 A \text{ cos A} - \text{cos}^3 A \text{sin A} \\[1em] \Rightarrow \text{sin A cos A} - \text{sin A cos A}(\text{sin}^2 A + \text{cos}^2 A)\end{aligned}$$

By formula,

sin² A + cos² A = 1.

$$\begin{aligned}\Rightarrow \text{sin A cos A} - \text{sin A cos A} \\[1em] \Rightarrow 0.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that sin A cos A - $\frac{\text{sin A cos (90° - A) cos A}}{\text{sec (90° - A)}} - \frac{\text{cos A sin (90° - A) sin A}}{\text{cosec (90° - A)}}$ = 0.

(i) In triangle ABC,

⇒ ∠A + ∠B + ∠C = 180° [By angle sum property of triangle]

⇒ ∠A + ∠B = 180° - ∠C .........(1)

Given equation,

sin $\frac{A + B}{2} = \text{cos} \frac{C}{2}$

Substituting value of (A + B) from (1) in L.H.S. of above equation :

$$\begin{aligned}\Rightarrow \text{sin} \frac{180° - C}{2} \\[1em] \Rightarrow \text{sin} \Big(90° - \frac{C}{2}\Big)\end{aligned}$$

By formula,

sin(90° - θ) = cos θ

$\therefore \text{cos} \frac{C}{2}$.

Since, L.H.S. = R.H.S.

Hence, proved that sin $\frac{A + B}{2} = \text{cos} \frac{C}{2}$.

(ii) In triangle ABC,

⇒ ∠A + ∠B + ∠C = 180° [By angle sum property of triangle]

⇒ ∠B + ∠C = 180° - ∠A .........(1)

Given equation,

tan $\frac{B + C}{2} = \text{cot} \frac{A}{2}$

Substituting value of (B + C) from (1) in L.H.S. of above equation :

$$\begin{aligned}\Rightarrow \text{tan} \frac{180° - A}{2} \\[1em] \Rightarrow \text{tan } \Big(90° - \frac{A}{2}\Big)\end{aligned}$$

By formula,

tan(90° - θ) = cot θ

$\Rightarrow \text{cot} \frac{A}{2}$.

Since, L.H.S. = R.H.S.

Hence, proved that tan $\frac{B + C}{2} = \text{cot} \frac{A}{2}$.

In triangle ABC,

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + 90° + ∠C = 180°

⇒ ∠A + ∠C = 90°

⇒ ∠A = 90° - ∠C.

Substituting value of A in $\frac{\text{sec A. cosec C - tan A. cot C}}{\text{sin B}}$ we get,

$\Rightarrow \frac{\text{sec (90° - C). cosec C - tan (90° - C). cot C}}{\text{sin 90°}}$

By formula,

tan (90° - C) = cot c, sec (90° - C) = cosec C and cosec² C - cot² C = 1.

$$\begin{aligned}\Rightarrow \frac{\text{cosec C. cosec C - cot C. cot C}}{1} \\[1em] \Rightarrow \text{cosec}^2 C - \text{cot}^2 C \\[1em] \Rightarrow 1.\end{aligned}$$

Hence, $\frac{\text{sec A. cosec C - tan A. cot C}}{\text{sin B}}$ = 1.

(i) Given,

⇒ sin (90° - 3A). cosec 42° = 1

⇒ sin (90° - 3A) × $\frac{1}{\text{sin 42°}}$ = 1

⇒ sin (90° - 3A) = sin 42°

⇒ 90° - 3A = 42°

⇒ 3A = 90° - 42°

⇒ 3A = 48°

⇒ A = 16°.

Hence, A = 16°.

(ii) Given,

⇒ cos (90° - A). sec 77° = 1

⇒ cos (90° - A) × $\frac{1}{\text{cos 77°}}$ = 1

⇒ cos (90° - A) = cos 77°

⇒ 90° - A = 77°

⇒ A = 90° - 77°

⇒ A = 13°.

Hence, A = 13°.

By formula,

cos (90° - θ) = sin θ.

Solving L.H.S. of the equation

$$\begin{aligned}\Rightarrow \frac{\text{cos(90° - θ) cos θ}}{\text{cot θ}} \\[1em] \Rightarrow \frac{\text{sin θ cos θ}}{\frac{\text{cos θ}}{\text{sin θ}}} \\[1em] \Rightarrow \text{sin}^2 θ \\[1em] \Rightarrow \text{1 - cos}^2 θ.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{cos(90° - θ) cos θ}}{\text{cot θ}}$ = 1 - cos² θ.

Solving,

⇒ sin² 34° + sin² 56° + 2 tan 18° tan 72° - cot² 30°

⇒ sin² 34° + sin² (90° - 34°) + 2 tan 18° tan (90° - 18°) - cot² 30°

By formula,

sin(90° - A) = cos A and tan(90° - A) = cot A

⇒ sin² 34° + cos² 34° + 2 tan 18° cot 18° - cot² 30°

By formula,

sin² θ + cos² θ = 1

⇒ 1 + 2 tan 18° $\times \frac{1}{\text{tan 18°}} - (\sqrt{3})^2$

⇒ 1 + 2 - 3

⇒ 3 - 3

⇒ 0.

Hence, sin² 34° + sin² 56° + 2 tan 18° tan 72° - cot² 30° = 0.

Solving,

⇒ cosec² 57° - tan² 33° + cos 44° cosec 46° - $\sqrt{2}$ cos 45° - tan² 60°

⇒ cosec² 57° - tan² (90° - 57°) + cos 44° cosec (90° - 44°) - $\sqrt{2} \times \frac{1}{\sqrt{2}} - (\sqrt{3})^2$

By formula,

tan(90° - A) = cot A and cosec(90° - A) = sec A

⇒ cosec² 57° - cot² 57° + cos 44° sec 44° - 1 - 3

By formula,

cosec² A - cot² A = 1

⇒ 1 + cos 44° $\times \frac{1}{\text{cos 44°}}$ - 1 - 3

⇒ 1 + 1 - 1 - 3

⇒ -2.

Hence, cosec² 57° - tan² 33° + cos 44° cosec 46° - $\sqrt{2}$ cos 45° - tan² 60° = -2.

Given,

⇒ 2 sin θ = $\sqrt{3}$

⇒ sin θ = $\frac{\sqrt{3}}{2}$

⇒ sin θ = sin 60°

⇒ θ = 60°

⇒ cos θ = cos 60°

⇒ 2 cos θ = 2 cos 60°

⇒ 2 cos θ = $2 \times \frac{1}{2}$

⇒ 2 cos θ = 1.

Hence, Option 4 is the correct option.

Given,

⇒ cos 62° 40' = x

⇒ cos (90° - 27° 20') = x

⇒ sin 27° 20' = x

Substituting value of sin 27° 20' in 1 + sin² 27° 20', we get :

⇒ 1 + sin² 27° 20' = 1 + x².

Hence, Option 1 is the correct option.

Given,

⇒ sin θ = 0.5

⇒ sin θ = $\frac{5}{10}$

⇒ sin θ = $\frac{1}{2}$

⇒ sin θ = sin 30°

⇒ θ = 30°.

Substituting value of θ in tan θ + cot θ, we get :

⇒ tan θ + cot θ

⇒ tan 30° + cot 30°

⇒ $\frac{1}{\sqrt{3}} + \sqrt{3}$

⇒ $\frac{1 + 3}{\sqrt{3}}$

⇒ $\frac{4}{\sqrt{3}}$

⇒ 4 : $\sqrt{3}$.

Hence, Option 3 is the correct option.

Given,

⇒ cos 63° 36' = 0.4446

⇒ cos (90° - 26° 24') = 0.4446

⇒ sin 26° 24' = 0.4446

Hence, Option 3 is the correct option.

(i) From table,

sin 21° = 0.3584

Hence, sin 21° = 0.3584

(ii) From table,

sin 34° 42' = 0.5693

Hence, sin 34° 42' = 0.5693

(iii) From table,

sin 47° 30' = 0.7373

Difference of 2' = 0.0004       [To add]

∴ sin 47° 32' = 0.7373 + 0.0004 = 0.7377

Hence, sin 47° 32' = 0.7377

(iv) From table,

sin 62° 54' = 0.8902

Difference of 3' = 0.0004       [To add]

∴ sin 62° 57' = 0.8902 + 0.0004 = 0.8906

Hence, sin 62° 57' = 0.8906

(i) From table,

cos 2° 6' = 0.9993

Difference of 2' = 0       [To add]

∴ cos 2° 4' = 0.9993 + 0 = 0.9993

Hence, 2° 4' = 0.9993

(ii) From table,

cos 8° 12' = 0.9898

Hence, 8° 12' = 0.9898

(iii) From table,

cos 26° 30' = 0.8949

Difference of 2' = 0.0003       [To subtract]

∴ cos 26° 32' = 0.8949 - 0.0003 = 0.8946

Hence, 26° 32' = 0.8946

(iv) From table,

cos 65° 42' = 0.4115

Difference of 1' = 0.0003       [To add]

∴ cos 65° 41' = 0.4115 + 0.0003 = 0.4118

Hence, 65° 41' = 0.4118

(i) From table,

tan 37° = 0.7536

Hence, tan 37° = 0.7536

(ii) From table,

tan 42° 18' = 0.9099

Hence, tan 42° 18' = 0.9099

(i) Let sin θ = 0.4848

From table,

sin 29° = 0.4848

⇒ θ = 29°.

Hence, θ = 29°.

(ii) Let sin θ = 0.3827

From graph,

sin 22° 30' = 0.3827

θ = 22° 30'.

Hence, θ = 22° 30'.

(i) Let cos θ = 0.9848

From graph,

cos 10° = 0.9848

θ = 10°.

Hence, θ = 10°.

(ii) Let cos θ = 0.9574

From graph,

cos 16° 48' = 0.9573

cos θ - cos 16° 48' = 0.9574 - 0.9573 = 0.0004

From table,

Difference of 1' = 0.0004 [To subtract]

∴ θ = 16° 48' - 1' = 16° 47'.

Hence, θ = 16° 47'.

(i) Let tan θ = 0.2419

From table,

tan 13° 36' = 0.2419

θ = 13° 36'.

Hence, θ = 13° 36'.

(ii) Let tan θ = 0.4741

From table,

tan 25° 18' = 0.4727

tan θ - tan 25° 18' = 0.4741 - 0.4727 = 0.0014

From table,

Difference of 4' = 0.0014

∴ θ = 25° 18' + 4' = 25° 22'.

Hence, θ = 25° 22'.

From figure,

cot α = $\frac{OA}{AB}$

cot β = $\frac{OA}{AC}$

cot γ = $\frac{OA}{AD}$

Since, AB < AC < AD

∴ cot γ < cot β < cot α

Hence, Option 1 is the correct option.

Given,

sin A = $\frac{4}{5}$

We need to find the value of:

$\sqrt{\frac{1 - \text{cos (90° - A)}}{1 + \text{cos (90° - A)}}}$

Solving,

$$\begin{aligned}\Rightarrow \sqrt{\frac{1 - \text{cos (90° - A)}}{1 + \text{cos (90° - A)}}} \\[1em] \Rightarrow \sqrt{\frac{1 - \text{sin A}}{1 + \text{sin A}}} \\[1em] \Rightarrow \sqrt{\frac{1 - \frac{4}{5}}{1 + \frac{4}{5}}} \\[1em] \Rightarrow \sqrt{\frac{\frac{5 - 4}{5}}{\frac{5 + 4}{5}}} \\[1em] \Rightarrow \sqrt{\frac{\frac{1}{5}}{\frac{9}{5}}} \\[1em] \Rightarrow \sqrt{\frac{1 \times 5}{5 \times 9}} \\[1em] \Rightarrow \sqrt{\frac{1}{9}} \\[1em] \Rightarrow \frac{1}{3}.\end{aligned}$$

Hence, Option 4 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow (1 + \text{tan}^2 A) \times \text{cos}^2 A - 1 \\[1em] \Rightarrow \Big(1 + \frac{\text{sin}^2 A}{\text{cos}^2 A}\Big) \times \text{cos}^2 A - 1 \\[1em] \Rightarrow \Big(\frac{\text{cos}^2 A + \text{sin}^2 A}{\text{cos}^2 A}\Big) \times \text{cos}^2 A - 1 \\[1em] \Rightarrow \frac{1}{\text{cos}^2 A} \times \text{cos}^2 A - 1 \\[1em] \Rightarrow 1 - 1 \\[1em] \Rightarrow 0.\end{aligned}$$

Hence, Option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{\text{sin A tan A}}{1 - \text{cos}^2 A} \\[1em] \Rightarrow \frac{\text{sin A tan A}}{\text{sin}^2 A} \\[1em] \Rightarrow \frac{\text{tan A}}{\text{sin A}} \\[1em] \Rightarrow \frac{\frac{\text{sin A}}{\text{cos A}}}{\text{sin A}} \\[1em] \Rightarrow \frac{\text{sin A}}{\text{sin A} \times \text{cos A}} \\[1em] \Rightarrow \frac{1}{\text{cos A}} \\[1em] \Rightarrow \text{sec A}.\end{aligned}$$

Hence, Option 2 is the correct option.

For acute angles,

Greater the angle, greater is the sine and tangent of the angle.

Since,

α > β

∴ tan α > tan β

∴ sin α > sin β.

Hence, Option 2 is the correct option.

Substituting value of M and N in M² + N², we get :

⇒ (x cos A + y sin A)² + (x sin A - y cos A)²

⇒ x² cos² A + y² sin² A + 2xy sin A cos A + x² sin² A + y² cos² A - 2xy sin A cos A

⇒ x² cos² A + y² sin² A + x² sin² A + y² cos² A

⇒ x² cos² A + x² sin² A + y² sin² A + y² cos² A

⇒ x² (sin² A + cos² A) + y² (sin² A + cos² A)

Substituting, sin² A + cos² A = 1, we get :

⇒ x² × 1 + y² × 1

⇒ x² + y².

Hence, Option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{\text{tan 60° - tan 30°}}{1 + \text{tan 60° tan 30°}} \\[1em] \Rightarrow \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3} \times \frac{1}{\sqrt{3}}} \\[1em] \Rightarrow \frac{\frac{3 - 1}{\sqrt{3}}}{1 + 1} \\[1em] \Rightarrow \frac{\frac{2}{\sqrt{3}}}{2} \\[1em] \Rightarrow \frac{2}{2\sqrt{3}} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} \\[1em] \Rightarrow \text{tan 30°}.\end{aligned}$$

Hence, Option 1 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow 3\frac{\text{sin 47°}}{\text{cos 43°}} - \frac{\text{tan 50°}}{\text{cot 40°}} - \text{2 sin}^2 45° \\[1em] \Rightarrow 3\frac{\text{sin 47°}}{\text{cos (90° - 47°)}} - \frac{\text{tan 50°}}{\text{cot (90° - 50°)}} - \text{2 sin}^2 45° \\[1em] \Rightarrow 3\frac{\text{sin 47°}}{\text{sin 47°}} - \frac{\text{tan 50°}}{\text{tan 50°}} - 2 \times \Big(\frac{1}{\sqrt{2}}\Big)^2 \\[1em] \Rightarrow 3 \times 1 - 1 - 2 \times \frac{1}{2} \\[1em] \Rightarrow 3 - 1 - 1 \\[1em] \Rightarrow 1.\end{aligned}$$

Hence, Option 3 is the correct option.

Solving,

⇒ sin 2x = 2 sin 45° cos 45°

⇒ sin 2x = $2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$

⇒ sin 2x = $2 \times \frac{1}{2}$

⇒ sin 2x = 1

⇒ sin 2x = sin 90°

⇒ 2x = 90°

⇒ x = $\frac{90°}{2}$ = 45°.

Hence, Option 1 is the correct option.

The valid trigonometric identity;

sec²θ = 1 + tan²θ

So, assertion (A) is true.

A trigonometric identity is an equation that holds true for all values of the variable (in this case, the angle θ) within its domain.

So, reason (R) is true.

∴ Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

Given,

⇒ (1 + tan² A)(1 - sin A)(1 + sin A)

⇒ (1 + tan² A)(1 - sin² A)

⇒ (1 + tan² A)cos² A

⇒ sec² A.cos² A

So, reason (R) are true.

⇒ $\frac{1}{\text{cos}^2 A}$ .cos² A

⇒ 1.

So, assertion (A) is false.

∴ A is false, R is true.

Hence, option 2 is the correct option.

Given, cos A = $\frac{\sqrt{3}}{2}$ and sin B = $\frac{1}{2}$

⇒ cos A = cos 30° and sin B = sin 30°

⇒ A = 30° and B = 30°

So, reason (R) is true.

tan(A + B) = tan(30° + 30°) = tan 60° = $\sqrt{3}$

Thus, tan (A + B) = $\sqrt{3}$

So, assertion (A) is true.

∴ Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

Given, x = (cosec A + cot A)(1 - cos A)

$$\begin{aligned}\Rightarrow x = \Big(\frac{1}{\text{sin A}} + \frac{\text{cos A}}{\text{sin A}}\Big)(1 - \text{cos A})\\[1em] = \Big(\frac{1 + \text{cos A}}{\text{sin A}}\Big)(1 - \text{cos A})\\[1em] = \frac{(1 + \text{cos A})(1 - \text{cos A})}{\text{sin A}}\\[1em] = \Big(\frac{1 - \text{cos}^2 A}{\text{sin A}}\Big)\\[1em] = \Big(\frac{\text{sin}^2 A}{\text{sin A}}\Big)\\[1em] = \text{sin A}.\end{aligned}$$

∴ Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

Given, $\frac{1 - \text{sin A}}{\text{cos A}}$ = sec A - tan A

Taking L.H.S.

$$\begin{aligned}\Rightarrow \frac{1 - \text{sin A}}{\text{cos A}}\\[1em] \Rightarrow \frac{1}{\text{cos A}} - \frac{\text{sin A}}{\text{cos A}}\\[1em] \Rightarrow \text{sec A} - \text{tan A}\end{aligned}$$

As, L.H.S. = R.H.S., so equation is true.

Now, $\frac{1 + \text{sin A}}{\text{cos A}}$ = sec A + tan A

Taking L.H.S.

$$\begin{aligned}\Rightarrow \frac{1 + \text{sin A}}{\text{cos A}}\\[1em] \Rightarrow \frac{1}{\text{cos A}} + \frac{\text{sin A}}{\text{cos A}}\\[1em] \Rightarrow \text{sec A} + \text{tan A}\end{aligned}$$

As, L.H.S. = R.H.S., so equation is true.

So, statement 1 is true.

Now, $\frac{1 - \text{sin A}}{\text{cos A}} - \frac{1 + \text{sin A}}{\text{cos A}}$ = 2sec A

Solving L.H.S.

$\Rightarrow \frac{1 - \text{sin A}}{\text{cos A}} - \frac{1 + \text{sin A}}{\text{cos A}}$

$\Rightarrow \frac{1}{\text{cos A}} - \frac{\text{sin A}}{\text{cos A}} - \Big(\frac{1}{\text{cos A}} + \frac{\text{sin A}}{\text{cos A}}\Big)$

⇒ (sec A - tan A) - (sec A + tan A)

⇒ sec A - tan A - sec A - tan A

⇒ -2tan A.

As, L.H.S. ≠ R.H.S.

So, statement 2 is false.

∴ Statement 1 is true and statement 2 is false.

Hence, option 3 is the correct option.

Given, $\text{cos}^2 θ + \frac{1}{1 + {\text{cot}^2 θ}}$ = x

$$\begin{aligned}\Rightarrow x = \text{cos}^2 θ + \frac{1}{1 + {\text{cot}^2 θ}}\\[1em] \Rightarrow x = \text{cos}^2 θ + \frac{1}{{\text{cosec}^2 θ}}\\[1em] \Rightarrow x = \text{cos}^2 θ + \text{sin}^2 θ\\[1em] \Rightarrow x = 1\end{aligned}$$

∴ Both the statements are true.

Hence, option 1 is the correct option.

Given, $\frac{\text{sin}^2 θ}{\text{cos}^2 θ} - \frac{1}{ \text{cos}^2 θ}$ = x

$$\begin{aligned}\Rightarrow x = \frac{\text{sin}^2 θ}{\text{cos}^2 θ} - \frac{1}{ \text{cos}^2 θ}\\[1em] \Rightarrow x = \frac{\text{sin}^2 θ - 1}{\text{cos}^2 θ}\\[1em] \Rightarrow x = \frac{\text{sin}^2 θ - (\text{sin}^2 θ + \text{cos}^2 θ)}{\text{cos}^2 θ} \\[1em] \Rightarrow x = \frac{\text{sin}^2 θ - \text{sin}^2 θ - \text{cos}^2 θ}{\text{cos}^2 θ}\\[1em] \Rightarrow x = \frac{- \text{cos}^2 θ}{\text{cos}^2 θ}\\[1em] \Rightarrow x = -1\end{aligned}$$

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Given, sin A = cos A

⇒ $\frac{\text{sin A}}{\text{cos A}}$ = 1

⇒ tan A = 1

⇒ tan A = tan 45°

⇒ A = 45°

So, statement 2 is true.

Substituting value of A in tan² A + cot² A + 2, we get :

⇒ tan² A + cot² A + 2 = tan² 45° + cot² 45° + 2

= (1)² + (1)² + 2

= 1 + 1 + 2

= 4.

So, statement 1 is true.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{1}{\text{cos A + sin A}} + \frac{1}{\text{cos A - sin A}} \\[1em] \Rightarrow \frac{\text{cos A - sin A + cos A + sin A}}{\text{cos}^2 A - \text{sin}^2 A} \\[1em] \Rightarrow \frac{\text{2 cos A}}{\text{cos}^2 A - \text{sin}^2 A}\end{aligned}$$

By formula,

sin² A = 1 - cos² A

$$\begin{aligned}\Rightarrow \frac{\text{2 cos A}}{\text{cos}^2 A - (1 - \text{cos}^2 A)} \\[1em] \Rightarrow \frac{\text{2 cos A}}{\text{2 cos}^2 A - 1}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{1}{\text{cos A + sin A}} + \frac{1}{\text{cos A - sin A}} = \frac{\text{2 cos A}}{\text{2 cos}^2 A - 1}$.

By formula,

sin² A = 1 - cos² A

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow 1 - \frac{1 - \text{cos}^2 A}{\text{1 + cos A}} \\[1em] \Rightarrow 1 - \frac{\text{(1 - cos A)(1 + cos A)}}{\text{1 + cos A}} \\[1em] \Rightarrow 1 - (\text{1 - cos A}) \\[1em] \Rightarrow 1 - 1 + \text{cos A} \\[1em] \Rightarrow \text{cos A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $1 - \frac{\text{sin}^2 A}{\text{1 + cos A}} = \text{cos A}$.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{cos A}}{\text{1 + sin A}} + \text{tan A}\\[1em] \Rightarrow \frac{\text{cos A}}{\text{1 + sin A}} + \frac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \frac{\text{cos}^2 A + \text{sin A(1 + sin A)}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \frac{\text{cos}^2 A + \text{sin A + sin}^2 A}{\text{cos A(1 + sin A)}}\end{aligned}$$

By formula,

cos² A + sin² A = 1

$$\begin{aligned}\Rightarrow \frac{\text{1 + sin A}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \frac{1}{\text{cos A}} \\[1em] \Rightarrow \text{sec A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{cos A}}{\text{1 + sin A}} + \text{tan A = sec A}$.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{sin A}}{\text{1 - cos A}} - \text{cot A} \\[1em] \Rightarrow \frac{\text{sin A}}{\text{1 - cos A}} - \frac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \frac{\text{sin}^2 A - \text{cos A(1 - cos A)}}{\text{sin A(1 - cos A)}} \\[1em] \Rightarrow \frac{\text{sin}^2 A - \text{cos A + cos}^2 A}{\text{sin A(1 - cos A)}}\end{aligned}$$

By formula,

sin² A + cos² A = 1

$$\begin{aligned}\Rightarrow \frac{\text{1 - cos A}}{\text{sin A(1 - cos A)}} \\[1em] \Rightarrow \frac{1}{\text{sin A}} \\[1em] \Rightarrow \text{cosec A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{sin A}}{\text{1 - cos A}} - \text{cot A = cosec A}$.

Multiplying numerator and denominator of L.H.S. of above equation by $\sqrt{1 + \text{cos A}}$ :

$$\begin{aligned}\Rightarrow \sqrt{\frac{1 - \text{cos A}}{\text{1 + cos A}}} \times \sqrt{\frac{1 + \text{cos A}}{1 + \text{cos A}}} \\[1em] \Rightarrow \sqrt{\frac{1 - \text{cos}^2 A}{(1 + \text{cos A})^2}}\end{aligned}$$

By formula,

1 - cos² A = sin² A

$$\begin{aligned}\Rightarrow \frac{\sqrt{\text{sin}^2 A}}{1 + \text{cos A}} \\[1em] \Rightarrow \frac{\text{sin A}}{\text{1 + cos A}}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\sqrt{\frac{1 - \text{cos A}}{\text{1 + cos A}}} = \frac{\text{sin A}}{\text{1 + cos A}}$.

By formula,

sec² - tan² A = 1

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{1 + \text{(sec A - tan A)}^2}{\text{cosec A (sec A - tan A)}} \\[1em] \Rightarrow \frac{\text{sec}^2 A - \text{tan}^2 A + \text{(sec A - tan A)}^2}{\text{cosec A (sec A - tan A)}} \\[1em] \Rightarrow \frac{\text{(sec A - tan A)(sec A + tan A) + (sec A - tan A)}^2}{\text{cosec A (sec A - tan A)}} \\[1em] \Rightarrow \frac{\text{(sec A - tan A)[sec A + tan A + sec A - tan A]}}{\text{cosec A (sec A - tan A)}} \\[1em] \Rightarrow \frac{\text{2 sec A}}{\text{cosec A}} \\[1em] \Rightarrow \frac{2 \times \frac{1}{\text{cos A}}}{\frac{1}{\text{sin A}}} \\[1em] \Rightarrow 2\frac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \text{2 tan A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{1 + \text{(sec A - tan A)}^2}{\text{cosec A (sec A - tan A)}} = \text{2 tan A}$.

By formula,

cosec² A - cot² A = 1

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{(cosec A - cot A)}^2 + 1}{\text{sec A (cosec A - cot A)}} \\[1em] \Rightarrow \frac{\text{(cosec A - cot A)}^2 + \text{cosec}^2 A - \text{cot}^2 A}{\text{sec A (cosec A - cot A)}} \\[1em] \Rightarrow \frac{\text{(cosec A - cot A)}^2 + \text{(cosec A - cot A)(cosec A + cot A)}}{\text{sec A (cosec A - cot A)}} \\[1em] \Rightarrow \frac{\text{(cosec A - cot A)(cosec A - cot A + cosec A + cot A)}}{\text{sec A (cosec A - cot A)}} \\[1em] \Rightarrow \frac{\text{2 cosec A}}{\text{sec A}} \\[1em] \Rightarrow \frac{\frac{2}{\text{sin A}}}{\frac{1}{\text{cos A}}} \\[1em] \Rightarrow \frac{\text{2 cos A}}{\text{sin A}} \\[1em] \Rightarrow \text{2 cot A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{(cosec A - cot A)}^2 + 1}{\text{sec A (cosec A - cot A)}} = \text{2 cot A}$.

Solving L.H.S. of above equation :

$$\begin{aligned}\Rightarrow \text{cot}^2 A \Big(\frac{\text{sec A - 1}}{\text{1 + sin A}}\times \frac{\text{sec A + 1}}{\text{sec A + 1}}\Big) + \text{sec}^2 A \Big(\frac{\text{sin A - 1}}{\text{1 + sec A}}\Big) \\[1em] \Rightarrow \text{cot}^2 A \Big(\frac{\text{sec}^2 A - 1}{\text{(1 + sin A)(sec A + 1)}}\Big) + \text{sec}^2 A \Big(\frac{\text{sin A - 1}}{\text{1 + sec A}}\Big)\end{aligned}$$

By formula,

sec² A - 1 = tan² A

$$\begin{aligned}\Rightarrow \frac{\text{cot}^2 A \text{ tan}^2 A}{\text{(1 + sin A)(sec A + 1)}} + \text{sec}^2 A \Big(\frac{\text{sin A - 1}}{\text{1 + sec A}}\Big) \\[1em] \Rightarrow \frac{\text{cot}^2 A \times \frac{1}{\text{ cot}^2 A}}{\text{(1 + sin A)(sec A + 1)}} + \text{sec}^2 A \Big(\frac{\text{sin A - 1}}{\text{1 + sec A}}\Big) \\[1em] \Rightarrow \frac{1}{\text{(1 + sin A)(sec A + 1)}} + \text{sec}^2 A \Big(\frac{\text{sin A - 1}}{\text{1 + sec A}}\Big) \\[1em] \Rightarrow \frac{1 + \text{sec}^2 A\text{(sin A - 1)(1 + sin A)}}{\text{(1 + sin A)(sec A + 1)}} \\[1em] \Rightarrow \frac{1 - \text{sec}^2 A\text{(1 - sin A)(1 + sin A)}}{\text{(1 + sin A)(sec A + 1)}} \\[1em] \Rightarrow \frac{1 - \text{sec}^2 A\text{(1 - sin}^2 A)}{\text{(1 + sin A)(sec A + 1)}} \\[1em] \Rightarrow \frac{1 - \text{sec}^2 A \times \text{cos}^2 A}{\text{(1 + sin A)(sec A + 1)}} \\[1em] \Rightarrow \frac{1 - \frac{1}{\text{cos}^2 A} \times \text{cos}^2 A}{\text{(1 + sin A)(sec A + 1)}} \\[1em] \Rightarrow \frac{1 - 1}{\text{(1 + sin A)(sec A + 1)}} \\[1em] \Rightarrow 0.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\text{cot}^2 A \Big(\frac{\text{sec A - 1}}{\text{1 + sin A}}\Big) + \text{sec}^2 A \Big(\frac{\text{sin A - 1}}{\text{1 + sec A}}\Big)$ = 0.

Solving L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \frac{(1 - \text{2 sin}^2 A)^2}{\text{cos}^4 A - \text{sin}^4 A} \\[1em] \Rightarrow \frac{(1 - \text{2 sin}^2 A)^2}{(\text{cos}^2 A - \text{sin}^2 A)(\text{cos}^2 A + \text{sin}^2 A)}\end{aligned}$$

By formula,

cos² A + sin² A = 1 and cos² A = 1 - sin² A.

$$\begin{aligned}\Rightarrow \frac{(1 - \text{2 sin}^2 A)^2}{(1 - \text{sin}^2 A - \text{sin}^2 A)} \\[1em] \Rightarrow \frac{(1 - \text{2 sin}^2 A)^2}{(1 - \text{2 sin}^2 A)}\\[1em] \Rightarrow \text{1 - 2 sin}^2 A \\[1em] \Rightarrow 1 - 2(\text{1 - cos}^2 A) \\[1em] \Rightarrow 1 - 2 + \text{2 cos}^2 A \\[1em] \Rightarrow \text{2 cos}^2 A - 1.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{(1 - \text{2 sin}^2 A)^2}{\text{cos}^4 A - \text{sin}^4 A}$ = 2 cos² A - 1.

Solving L.H.S. of the above equation :

⇒ sec⁴ A (1 - sin⁴ A) - 2 tan² A

⇒ sec⁴ A (1 - sin² A)(1 + sin² A) - 2 tan² A

By formula,

1 - sin² A = cos² A

⇒ sec⁴ A cos² A (1 + sin² A) - 2 tan² A

⇒ sec⁴ A $\times \frac{1}{\text{sec}^2 A}$ (1 + sin² A) - 2 tan² A

⇒ sec² A (1 + sin² A) - 2 tan² A

⇒ sec² A + sec² A sin² A - 2 tan² A

⇒ sec² A + $\frac{1}{\text{cos}^2 A} \times$ sin² A - 2 tan² A

⇒ sec² A + tan² A - 2 tan² A

⇒ sec² A - tan² A

⇒ 1.

Since, L.H.S. = R.H.S.

Hence, proved that sec⁴ A (1 - sin⁴ A) - 2 tan² A = 1.

Solving L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \Big(1 + \frac{\text{sin A}}{\text{cos A}} + \frac{1}{\text{cos A}}\Big)\Big(1 + \frac{\text{cos A}}{\text{sin A}} - \frac{1}{\text{sin A}}\Big) \\[1em] \Rightarrow \Big(\frac{\text{cos A + sin A + 1}}{\text{cos A}}\Big)\Big(\frac{\text{sin A + cos A - 1}}{\text{sin A}}\Big) \\[1em] \Rightarrow \frac{\text{(sin A + cos A)}^2 - (1)^2}{\text{sin A cos A}} \\[1em] \Rightarrow \frac{\text{sin}^2 A + \text{cos}^2 A + \text{2 sin A cos A} - 1}{\text{sin A cos A}}\end{aligned}$$

By formula,

sin² A + cos² A = 1

$$\begin{aligned}\Rightarrow \frac{\text{1 - 1 + 2 sin A cos A}}{\text{sin A cos A}}\\[1em] \Rightarrow \frac{\text{2 sin A cos A}}{\text{sin A cos A}} \\[1em] \Rightarrow 2.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that (1 + tan A + sec A)(1 + cot A - cosec A) = 2.

Substituting value of x and y in L.H.S. of above equation :

$$\begin{aligned}\Rightarrow \frac{a^2}{a^2 \text{ cos}^2 θ} - \frac{b^2}{b^2 \text{ cot}^2 θ} \\[1em] \Rightarrow \frac{1}{\text{cos}^2 θ} - \frac{1}{\text{cot}^2 θ} \\[1em] \Rightarrow \frac{1}{\text{cos}^2 θ} - \frac{1}{\frac{\text{cos}^2 θ}{\text{sin}^2 θ}} \\[1em] \Rightarrow \frac{1}{\text{cos}^2 θ} - \frac{\text{sin}^2 θ}{\text{cos}^2 θ} \\[1em] \Rightarrow \frac{1 - \text{sin}^2 θ}{\text{cos}^2 θ} \\[1em] \Rightarrow \frac{\text{cos}^2 θ}{\text{cos}^2 θ} \\[1em] \Rightarrow 1.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{a^2}{x^2} - \frac{b^2}{y^2}$ = 1.

Substituting value of p in R.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \frac{\text{(sec A + tan A)}^2 - 1}{\text{(sec A + tan A)}^2 + 1} \\[1em] \Rightarrow \frac{\text{sec}^2 A + \text{tan}^2 A + \text{2 sec A tan A} - 1}{\text{sec}^2 A + \text{tan}^2 A + \text{2 sec A tan A} + 1}\end{aligned}$$

By formula,

sec² A - 1 = tan² A and sec² A = 1 + tan² A

$$\begin{aligned}\Rightarrow \frac{\text{sec}^2 A - 1 + \text{tan}^2 A + \text{2 sec A tan A}}{\text{sec}^2 A + \text{tan}^2 A + 1 + \text{2 sec A tan A}} \\[1em] \Rightarrow \frac{\text{tan}^2 A + \text{tan}^2 A + \text{2 sec A tan A}}{\text{sec}^2 A + \text{sec}^2 A + \text{2 sec A tan A}} \\[1em] \Rightarrow \frac{\text{2 tan}^2 A + \text{2 sec A tan A}}{\text{2 sec}^2 A + \text{2 sec A tan A}} \\[1em] \Rightarrow \frac{2\text{tan A}(\text{tan A + sec A})}{\text{2 sec A(sec A + tan A)}} \\[1em] \Rightarrow \frac{\text{tan A}}{\text{sec A}} \\[1em] \Rightarrow \frac{\frac{\text{sin A}}{\text{cos A}}}{\frac{1}{\text{cos A}}} \\[1em] \Rightarrow \frac{\text{sin A}}{\text{cos A}} \times \text{cos A} \\[1em] \Rightarrow \text{sin A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that sin A = $\frac{p^2 - 1}{p^2 + 1}$.

Given equation : tan (55° + x) = cot (35° - x)

Solving L.H.S.

⇒ tan (55° + x)

⇒ tan [90° - (35° - x)]

By formula,

tan (90° - θ) = cot θ

⇒ cot (35° - x)

Since, L.H.S. = R.H.S.

Hence, proved that tan (55° + x) = cot (35° - x).

Given equation : sec (70° - θ) = cosec (20° + θ)

Solving L.H.S.

⇒ sec (70° - θ)

⇒ sec [90° - (20° + θ)]

By formula,

sec (90° - θ) = cosec θ

⇒ cosec (20° + θ)

Since, L.H.S. = R.H.S.

Hence, proved that sec (70° - θ) = cosec (20° + θ).

Given equation : sin(28° + A) = cos(62° - A)

Solving L.H.S.

⇒ sin(28° + A)

⇒ sin [90° - (62° - A)]

By formula,

sin (90° - θ) = cos θ

⇒ cos (62° - A)

Since, L.H.S. = R.H.S.

Hence, proved that sin(28° + A) = cos(62° - A).

By formula,

cos (90° - A) = sin A

Solving L.H.S. of above equation,

$$\begin{aligned}\Rightarrow \frac{1}{\text{1 + sin A}} + \frac{1}{1 - \text{sin A}} \\[1em] \Rightarrow \frac{\text{1 - sin A + 1 + sin A}}{\text{(1 + sin A)(1 - sin A)}} \\[1em] \Rightarrow \frac{2}{1 - \text{sin}^2 A} \\[1em] \Rightarrow \frac{2}{\text{cos}^2 A} \\[1em] \Rightarrow 2\text{ sec}^2 A \\[1em] \Rightarrow 2\text{ cosec}^2 (90° - A) \space [\because \text{ sec A} = \text{ cosec(90° - A)}]\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{1}{\text{1 + cos (90° - A)}} + \frac{1}{\text{1 - cos (90° - A)}}$ = 2 cosec² (90° - A).

Given,

A + B = 90°

B = 90° - A and A = 90° - B.

(i) Substituting value of B in L.H.S. of equation :

$$\begin{aligned}\Rightarrow \text{cot (90° - A) + cos (90° - A)} \\[1em] \Rightarrow \text{tan A + sin A} \\[1em] \Rightarrow \frac{\text{sin A}}{\text{cos A}} + \text{sin A} \\[1em] \Rightarrow \frac{\text{sin A + sin A cos A}}{\text{cos A}} \\[1em] \Rightarrow \frac{\text{sin A(1 + cos A)}}{\text{cos A}}\\[1em] \Rightarrow \text{sin A sec A (1 + cos A)} \\[1em] \Rightarrow \text{sin (90° - B) sec A [1 + cos (90° - B)]}\end{aligned}$$

By formula,

sin (90° - B) = cos B and cos (90° - B) = sin B

⇒ cos B sec A (1 + sin B).

Since, L.H.S. = R.H.S.

Hence, proved that cot B + cos B = sec A cos B (1 + sin B).

(ii) Substituting value of B in L.H.S. of equation :

⇒ cot A cot (90° - A) - sin A cos (90° - A) - cos A sin (90° - A)

By formula,

cot (90° - A) = tan A, cos (90° - A) = sin A and sin (90° - A) = cos A

⇒ cot A tan A - sin A sin A - cos A cos A

⇒ cot A $\times \frac{1}{\text{cot A}}$ - (sin² A + cos² A)

⇒ 1 - 1

⇒ 0.

Since, L.H.S. = R.H.S.

Hence, proved that cot A cot B - sin A cos B - cos A sin B = 0.

(iii) Substituting value of B in L.H.S. of equation :

⇒ cosec² A + cosec² (90° - A)

By formula,

cosec (90° - A) = sec A

⇒ $\frac{1}{\text{sin}^2 A}$ + sec² A

⇒ $\frac{1}{\text{sin}^2 A} + \frac{1}{\text{cos}^2 A}$

⇒ $\frac{\text{cos}^2 A + \text{sin}^2 A}{\text{sin}^2 A \text{cos}^2 A}$

By formula,

sin² A + cos² A = 1 and sec (90° - A) = cosec A

⇒ $\frac{1}{\text{sin}^2 A \text{cos}^2 A}$

⇒ cosec² A sec² A

⇒ cosec² A sec² (90° - B)

⇒ cosec² A cosec² B

Since, L.H.S. = R.H.S.

Hence, proved that cosec² A + cosec² B = cosec² A cosec² B.

(iv) Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{sin A + sin B}}{\text{sin A - sin B}} + \frac{\text{cos B - cos A}}{\text{cos B + cos A}} \\[1em] \Rightarrow \frac{\text{sin A + sin B}}{\text{sin A - sin B}} + \frac{\text{cos (90° - A) - cos (90° - B)}}{\text{cos (90° - A) + cos (90° - B)}}\end{aligned}$$

By formula,

cos (90° - θ) = sin θ and sin (90° - θ) = cos θ.

$$\begin{aligned}\Rightarrow \frac{\text{sin A + sin B}}{\text{sin A - sin B}} + \frac{\text{sin A - sin B}}{\text{sin A + sin B}} \\[1em] \Rightarrow \frac{\text{(sin A + sin B)}^2 + \text{(sin A - sin B)}^2}{\text{(sin A - sin B)(sin A + sin B)}} \\[1em] \Rightarrow \frac{\text{sin}^2 A + \text{sin}^2 B + \text{2 sin A sin B + sin}^2 A + \text{sin}^2 B - \text{2 sin A sin B}}{\text{sin}^2 A - \text{sin}^2 B} \\[1em] \Rightarrow 2 \frac{\text{sin}^2 A + \text{sin}^2 B}{\text{sin}^2 A - \text{sin}^2 B} \\[1em] \Rightarrow 2 \frac{\text{sin}^2 A + \text{sin}^2 (90° - A)}{\text{sin}^2 A - \text{sin}^2 (90° - A)} \\[1em] \Rightarrow 2 \frac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin}^2 A - \text{cos}^2 A} \\[1em] \Rightarrow \frac{2 \times 1}{\text{sin}^2 A - \text{cos}^2 A} \\[1em] \Rightarrow \frac{2}{\text{sin}^2 A - \text{cos}^2 A}\end{aligned}$$

By formula,

cos² A = 1 - sin² A

$$\begin{aligned}\Rightarrow \frac{2}{\text{sin}^2 A - (1 - \text{sin}^2 A)} \\[1em] \Rightarrow \frac{2}{\text{sin}^2 A + \text{sin}^2 A - 1} \\[1em] \Rightarrow \frac{2}{\text{2 sin}^2 A - 1}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{sin A + sin B}}{\text{sin A - sin B}} + \frac{\text{cos B - cos A}}{\text{cos B + cos A}} = \frac{2}{\text{2 sin}^2 A - 1}$.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{1}{\text{sin A - cos A}} - \frac{1}{\text{sin A + cos A}} \\[1em] \Rightarrow \frac{\text{sin A + cos A - (sin A - cos A)}}{\text{(sin A - cos A)(sin A + cos A)}} \\[1em] \Rightarrow \frac{\text{sin A - sin A + cos A + cos A}}{\text{sin}^2 A - \text{cos}^2 A} \\[1em] \Rightarrow \frac{\text{2 cos A}}{\text{sin}^2 A - \text{cos}^2 A}\end{aligned}$$

By formula,

cos² A = 1 - sin² A

$$\begin{aligned}\Rightarrow \frac{\text{2 cos A}}{\text{sin}^2 A - (1 - \text{sin}^2 A)} \\[1em] \Rightarrow \frac{\text{2 cos A}}{\text{sin}^2 A + \text{sin}^2 A - 1} \\[1em] \Rightarrow \frac{\text{2 cos A}}{\text{2 sin}^2 A - 1}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{1}{\text{sin A - cos A}} - \frac{1}{\text{sin A + cos A}} = \frac{\text{2 cos A}}{\text{2 sin}^2 A - 1}$.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{cot}^2 A}{\text{cosec A - 1}} - 1 \\[1em] \Rightarrow \frac{\text{cot}^2 A - \text{(cosec A - 1)}}{\text{cosec A - 1}} \\[1em] \Rightarrow \frac{\text{cot}^2 A + 1 - \text{cosec A}}{\text{cosec A - 1}}\end{aligned}$$

By formula,

cot² A + 1 = cosec² A

$$\begin{aligned}\Rightarrow \frac{\text{cosec}^2 A - \text{cosec A}}{\text{cosec A - 1}} \\[1em] \Rightarrow \frac{\text{cosec A(cosec A - 1)}}{\text{cosec A - 1}} \\[1em] \Rightarrow \text{cosec A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{cot}^2 A}{\text{cosec A - 1}} - 1$ = cosec A.

Solving R.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \text{sec A - tan A} \\[1em] \Rightarrow \frac{1}{\text{cos A}} - \frac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \frac{\text{1 - sin A}}{\text{cos A}} \\[1em] \Rightarrow \frac{\text{1 - sin A}}{\text{cos A}} \times \frac{\text{1 + sin A}}{\text{1 + sin A}} \\[1em] \Rightarrow \frac{\text{1 - sin}^2 A}{\text{cos A(1+ sin A)}} \\[1em] \Rightarrow \frac{\text{1 - sin}^2 A}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \frac{\text{cos}^2 A}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \frac{\text{cos A}}{\text{1 + sin A}}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{cos A}}{\text{1 + sin A}}$ = sec A - tan A.

Solving L.H.S. of the above equation :

⇒ cos A(1 + cot A) + sin A(1 + tan A)

⇒ cos A + cos A cot A + sin A + sin A tan A

⇒ cos A + cos A $\times \frac{\text{cos A}}{\text{sin A}} + \text{sin A} + \text{sin A} \times \frac{\text{sin A}}{\text{cos A}}$

⇒ cos A + $\frac{\text{sin}^2 A}{\text{cos A}} + \text{ sin A } + \frac{\text{cos}^2 A}{\text{sin A}}$

⇒ $\frac{\text{cos}^2 A + \text{sin}^2 A}{\text{cos A}} + \frac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A}}$

By formula,

sin² A + cos² A = 1.

⇒ $\frac{1}{\text{cos A}} + \frac{1}{\text{sin A}}$

⇒ sec A + cosec A.

Since, L.H.S. = R.H.S.

Hence, proved that cos A(1 + cot A) + sin A(1 + tan A) = sec A + cosec A.

Solving L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \sqrt{\Big(\frac{1}{\text{cos A}}\Big)^2 + \Big(\frac{1}{\text{sin A}}\Big)^2} \\[1em] \Rightarrow \sqrt{\frac{1}{\text{cos}^2 A} + \frac{1}{\text{sin}^2 A}} \\[1em] \Rightarrow \sqrt{\frac{\text{sin}^2 A + \text{cos}^2 A}{\text{cos}^2 A\text{ sin}^2 A}}\end{aligned}$$

By formula,

sin² A + cos² A = 1

$$\begin{aligned}\Rightarrow \sqrt{\frac{1}{\text{cos}^2 A\text{ sin}^2 A}} \\[1em] \Rightarrow \frac{1}{\text{sin A}\text{ cos A}}.\end{aligned}$$

Solving R.H.S. of the equation :

$$\begin{aligned}\Rightarrow \text{tan A + cot A} \\[1em] \Rightarrow \frac{\text{sin A}}{\text{cos A}} + \frac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \frac{\text{sin}^2 A + \text{cos}^2 A}{\text{cos A sin A}} \\[1em] \Rightarrow \frac{1}{\text{sin A cos A}}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\sqrt{\text{sec}^2 A + \text{cosec}^2 A}$ = tan A + cot A.

Solving L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \text{sin A sec A + sin A cosec A + cos A sec A + cos A cosec A} \\[1em] \Rightarrow \text{sin A} \times \frac{1}{\text{cos A}} + \text{sin A} \times \frac{1}{\text{sin A}} + \text{cos A} \times \frac{1}{\text{cos A}} + \text{cos A} \times \frac{1}{\text{sin A}} \\[1em] \Rightarrow \frac{\text{sin A}}{\text{cos A}} + 1 + 1 + \frac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow 2 + \frac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}} \\[1em] \Rightarrow 2 + \frac{1}{\text{sin A cos A}} \\[1em] \Rightarrow 2 + \text{cosec A sec A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that (sin A + cos A)(sec A + cosec A) = 2 + sec A cosec A.

Solving L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \text{(tan A + cot A)(cosec A - sin A)(sec A - cos A)} \\[1em] \Rightarrow \Big(\frac{\text{sin A}}{\text{cos A}} + \frac{\text{cos A}}{\text{sin A}}\Big)\Big(\frac{1}{\text{sin A}} - \text{sin A}\Big)\Big(\frac{1}{\text{cos A}} - \text{cos A}\Big) \\[1em] \Rightarrow \Big(\frac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}}\Big)\Big(\frac{1 - \text{sin}^2 A}{\text{sin A}}\Big)\Big(\frac{1 - \text{cos}^2 A}{\text{cos A}}\Big)\end{aligned}$$

By formula,

sin² A + cos² A = 1, 1 - sin² A = cos² A and 1 - cos² A = sin² A.

$$\begin{aligned}\Rightarrow \frac{1}{\text{sin A cos A}} \times \frac{\text{cos}^2 A}{\text{sin A}} \times \frac{\text{sin}^2 A}{\text{cos A}} \\[1em] \Rightarrow \frac{\text{sin}^2 A \text{cos}^2 A}{\text{sin}^2 A \text{cos}^2 A} \\[1em] \Rightarrow 1.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that (tan A + cot A)(cosec A - sin A)(sec A - cos A) = 1.

Solving,

$$\begin{aligned}\Rightarrow \text{cot}^2 A - \text{cot}^2 B \\[1em] \Rightarrow \frac{\text{cos}^2 A}{\text{sin}^2 A} - \frac{\text{cos}^2 B}{\text{sin}^2 B} \\[1em] \Rightarrow \frac{\text{cos}^2 A \text{ sin}^2 B - \text{cos}^2 B \text{ sin}^2 A}{\text{sin}^2 A \text{ sin}^2 B} \\[1em]\end{aligned}$$

By formula,

sin² θ = 1 - cos² θ

$$\begin{aligned}\Rightarrow \frac{\text{cos}^2 A (1 - \text{ cos}^2 B) - \text{cos}^2 B (1 - \text{ cos}^2 A)}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow \frac{\text{cos}^2 A - \text{cos}^2 A \text{cos}^2 B - \text{cos}^2 B + \text{cos}^2 A \text{cos}^2 B}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow \frac{\text{cos}^2 A - \text{cos}^2 B}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow \frac{\text{1 - sin}^2 A - (1 - \text{sin}^2 B)}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow \frac{1 - 1 - \text{sin}^2 A + \text{sin}^2 B}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow \frac{-\text{sin}^2 A + \text{sin}^2 B}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow -\frac{\text{sin}^2 A}{\text{sin}^2 A \text{ sin}^2 B} + \frac{\text{sin}^2 B}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow -\frac{1}{\text{sin}^2 B} + \frac{1}{\text{sin}^2 A} \\[1em] \Rightarrow -\text{cosec}^2 B + \text{cosec}^2 A \\[1em] \Rightarrow \text{cosec}^2 A - \text{cosec}^2 B.\end{aligned}$$

Hence, proved that cot² A - cot² B = $\frac{\text{cos}^2 \text{A} - \text{cos}^2 \text{B}}{\text{sin}^2 \text{A} \text{ sin}^2 \text{B}}$ = cosec² A - cosec² B.

Solving L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \frac{\frac{1}{\text{tan A}} - 1}{\text{2 - (1 + tan}^2 A)} \\[1em] \Rightarrow \frac{\frac{\text{1 - tan A}}{\text{tan A}}}{2 - 1 - \text{tan}^2 A} \\[1em] \Rightarrow \frac{\text{1 - tan A}}{\text{tan A(1 - tan}^2 A)} \\[1em] \Rightarrow \frac{\text{1 - tan A}}{\text{tan A(1 - tan A)(1 + tan A)}} \\[1em] \Rightarrow \frac{1}{\text{tan A(1 + tan A)}} \\[1em] \Rightarrow \frac{1}{\frac{1}{\text{cot A}} \text{(1 + tan A)}} \\[1em] \Rightarrow \frac{\text{cot A}}{\text{1 + tan A}}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{cot A - 1}}{\text{2 - sec}^2 A} = \frac{\text{cot A}}{\text{1 + tan A}}$.

Given,

⇒ 4 cos² A - 3 = 0

⇒ 4 cos² A = 3

⇒ cos² A = $\frac{3}{4}$

⇒ cos A = $\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

⇒ cos A = cos 30°

⇒ A = 30°.

(i) To prove:

sin 3A = 3 sin A - 4 sin³ A

Solving L.H.S. of the equation :

⇒ sin 3A = sin 3(30°)

= sin 90° = 1.

Solving R.H.S. of the equation :

⇒ 3 sin A - 4 sin³ A

⇒ 3 sin 30° - 4 sin³ 30°

⇒ 3 $\times \frac{1}{2} - 4 \times \Big(\frac{1}{2}\Big)^3$

⇒ $\frac{3}{2} - 4 \times \frac{1}{8}$

⇒ $\frac{3}{2} - \frac{1}{2}$

⇒ $\frac{2}{2}$

⇒ 1.

Since, L.H.S. = R.H.S.

Hence, proved that sin 3A = 3 sin A - 4 sin³ A.

(ii) To prove:

cos 3A = 4 cos³ A - 3 cos A

Solving L.H.S.

⇒ cos 3A = cos 3(30°) = cos 90° = 0.

Solving R.H.S.

⇒ 4 cos³ A - 3 cos A

⇒ 4 cos³ 30° - 3 cos 30°

⇒ 4 $\times \Big(\frac{\sqrt{3}}{2}\Big)^3 - 3 \times \frac{\sqrt{3}}{2}$

⇒ 4 $\times \frac{3\sqrt{3}}{8} - \frac{3\sqrt{3}}{2}$

⇒ $\frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2}$

⇒ 0.

Since, L.H.S. = R.H.S.

Hence, proved that cos 3A = 4 cos³ A - 3 cos A

(i) Solving,

⇒ 2 cos² A - 1 = 0

⇒ 2 cos² A = 1

⇒ cos² A = $\frac{1}{2}$

⇒ cos A = $\sqrt{\frac{1}{2}}$

⇒ cos A = $\frac{1}{\sqrt{2}}$

⇒ cos A = cos 45°

⇒ A = 45°.

Hence, A = 45°.

(ii) Solving,

⇒ sin 3A - 1 = 0

⇒ sin 3A = 1

⇒ sin 3A = sin 90°

⇒ 3A = 90°

⇒ A = 30°.

Hence, A = 30°.

(iii) Solving,

⇒ 4 sin² A - 3 = 0

⇒ 4 sin² A = 3

⇒ sin² A = $\frac{3}{4}$

⇒ sin A = $\sqrt{\frac{3}{4}}$

⇒ sin A = $\frac{\sqrt{3}}{2}$

⇒ sin A = sin 60°

⇒ A = 60°.

Hence, A = 60°.

(iv) Solving,

⇒ cos² A - cos A = 0

⇒ cos A(cos A - 1) = 0

⇒ cos A = 0 or cos A - 1 = 0

⇒ cos A = 0 or cos A = 1

⇒ cos A = cos 90° or cos A = cos 0°

⇒ A = 90° or A = 0°.

Hence, A = 0° or 90°.

(v) Solving,

⇒ 2 cos² A + cos A - 1 = 0

⇒ 2 cos² A + 2 cos A - cos A - 1 = 0

⇒ 2 cos A(cos A + 1) - 1(cos A + 1) = 0

⇒ (2 cos A - 1)(cos A + 1) = 0

⇒ 2 cos A = 1 or cos A = -1

⇒ cos A = $\frac{1}{2}$ or cos A = -1

Since, cos A cannot be negative in the range 0° ≤ A ≤ 90°.

∴ cos A = $\frac{1}{2}$

⇒ cos A = cos 60°

⇒ A = 60°

Hence, A = 60°.

(i) Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{cos A}}{\text{1 - sin A}} + \frac{\text{cos A}}{\text{1 + sin A}} = 4 \\[1em] \Rightarrow \frac{\text{cos A(1 + sin A) + cos A(1 - sin A)}}{\text{(1 + sin A)(1 - sin A)}} \\[1em] \Rightarrow \frac{\text{cos A + cos A sin A + cos A - cos A sin A}}{\text{1 - sin}^2 A}\end{aligned}$$

By formula,

1 - sin² A = cos² A

$$\begin{aligned}\Rightarrow \frac{\text{2 cos A}}{\text{cos}^2 A} \\[1em] \Rightarrow \frac{2}{\text{cos A}} \\[1em] \Rightarrow 2\text{ sec A}.\end{aligned}$$

Given, R.H.S. = 4

∴ 2 sec A = 4

⇒ sec A = 2

⇒ sec A = sec 60°

⇒ A = 60°.

Hence, A = 60°.

(ii) Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{sin A}}{\text{sec A - 1}} + \frac{\text{sin A}}{\text{sec A + 1}} = 2 \\[1em] \Rightarrow \frac{\text{sin A(sec A + 1) + sin A(sec A - 1)}}{\text{(sec A - 1)(sec A + 1)}} \\[1em] \Rightarrow \frac{\text{sin A sec A + sin A + sin A sec A - sin A}}{\text{sec}^2 A - 1} \\[1em] \Rightarrow \frac{\text{2 sin A sec A}}{\text{sec}^2 A - 1}\end{aligned}$$

By formula,

sec² A - 1 = tan² A

$$\begin{aligned}\Rightarrow \frac{\text{2 sin A} \times \frac{1}{\text{cos A}}}{\text{tan}^2 A} \\[1em] \Rightarrow \frac{\text{2 tan A}}{\text{tan}^2 A} \\[1em] \Rightarrow \frac{2}{\text{tan A}} \\[1em] \Rightarrow \text{2 cot A}.\end{aligned}$$

Given, R.H.S. = 2

∴ 2 cot A = 2

⇒ cot A = 1

⇒ cot A = cot 45°

⇒ A = 45°.

Hence, A = 45°.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \Big(\frac{1}{\text{sin A}} - \text{sin A}\Big) \Big(\frac{1}{\text{cos A}} - \text{cos A}\Big)\text{sec}^2 \text{A} \\[1em] \Rightarrow \Big(\frac{\text{1 - sin}^2 A}{\text{sin A}}\Big) \times \Big(\frac{\text{1 - cos}^2 A}{\text{cos A}}\Big) \times \text{sec}^2 A\end{aligned}$$

By formula,

1 - sin² A = cos² A and 1 - cos² A = sin² A

$$\begin{aligned}\Rightarrow \frac{\text{cos}^2 A}{\text{sin A}} \times \frac{\text{sin}^2 A}{\text{cos A}} \times \frac{1}{\text{cos}^2 A} \\[1em] \Rightarrow \frac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \text{tan A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that (cosec A - sin A)(sec A - cos A) sec² A = tan A

Given equation,

⇒ (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.

Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \text{(sin θ + cos θ)(tan θ + cot θ)} \\[1em] \Rightarrow \text{(sin θ + cos θ)}\Big(\frac{\text{sin θ}}{\text{cos θ}} + \frac{\text{cos θ}}{\text{sin θ}}\Big) \\[1em] \Rightarrow \text{(sin θ + cos θ)}\Big(\frac{\text{sin}^2 θ + \text{cos}^2 θ}{\text{cos θ sin θ}}\Big)\end{aligned}$$

By formula,

sin² θ + cos² θ = 1.

$$\begin{aligned}\Rightarrow \text{(sin θ + cos θ)} \times \frac{1}{\text{cos θ sin θ}} \\[1em] \Rightarrow \frac{\text{sin θ}}{\text{cos θ sin θ}} + \frac{\text{cos θ}}{\text{cos θ sin θ}} \\[1em] \Rightarrow \frac{1}{\text{cos θ}} + \frac{1}{\text{sin θ}} \\[1em] \Rightarrow \text{sec θ + cosec θ}.\end{aligned}$$

Hence, proved that (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.

Solving,

⇒ sin² 28° + sin² 62° + tan² 38° - cot² 52° + $\frac{1}{4} \text{sec}^2 \space 30°$

⇒ sin² 28° + sin² (90° - 28°) + tan² (90° - 52°) - cot² 52° + $\frac{1}{4} \times \Big(\frac{2}{\sqrt{3}}\Big)^2$

By formula,

sin(90° - θ) = cos θ and tan(90° - θ) = cot θ

⇒ sin² 28° + cos² 28° + cot² 52° - cot² 52° + $\frac{1}{4} \times \frac{4}{3}$

By formula,

sin² θ + cos² θ = 1

⇒ 1 + $\frac{1}{3}$

⇒ $1\frac{1}{3}$.

Hence, sin² 28° + sin² 62° + tan² 38° - cot² 52° + $\frac{1}{4} \text{sec}^2 30° = 1\frac{1}{3}$.

(i) Solving L.H.S. of the equation :

⇒ (1 + cot² A)(1 - cos A)(1 + cos A)

⇒ (1 + cot² A)(1 - cos² A)

By formula,

1 + cot² A = cosec² A and 1 - cos² A = sin² A.

$$\begin{aligned}\Rightarrow \text{cosec}^2 A \times \text{sin}^2 A \\[1em] \Rightarrow \frac{1}{\text{sin}^2 A} \times \text{sin}^2 A \\[1em] \Rightarrow 1.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that (1 + cot² A)(1 - cos A)(1 + cos A) = 1.

(ii) Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{sin}^4 A + \text{cos}^4 A}{\text{sin}^2 A. \text{cos}^2 A} \\[1em] \Rightarrow \frac{\text{sin}^4 A}{\text{sin}^2 A. \text{cos}^2 A} + \frac{\text{cos}^4 A}{\text{sin}^2 A. \text{cos}^2 A} \\[1em] \Rightarrow \frac{\text{sin}^2 A}{\text{cos}^2 A} + \frac{\text{cos}^2 A}{\text{sin}^2 A} \\[1em] \Rightarrow \text{tan}^2 A + \text{cot}^2 A.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{sin}^4 A + \text{cos}^4 A}{\text{sin}^2 A. \text{cos}^2 A}$ = tan² A + cot² A.

(iii) Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \text{cosec A(1 - cos A)(cosec A + cot A)} \\[1em] \Rightarrow \frac{1}{\text{sin A}}(1 - \text{cos A})\Big(\frac{1}{\text{sin A}} + \frac{\text{cos A}}{\text{sin A}}\Big) \\[1em] \Rightarrow \frac{1}{\text{sin A}}(1 - \text{cos A}) \times \frac{\text{1 + cos A}}{\text{sin A}} \\[1em] \Rightarrow \frac{\text{(1 - cos A)(1 + cos A)}}{\text{sin}^2 A} \\[1em] \Rightarrow \frac{1 - \text{cos}^2 A}{\text{sin}^2 A}\end{aligned}$$

By formula,

1 - cos² A = sin² A.

$$\begin{aligned}\Rightarrow \frac{\text{sin}^2 A}{\text{sin}^2 A} \\[1em] \Rightarrow 1.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that cosec A(1 - cos A)(cosec A + cot A) = 1.

(iv) Solving L.H.S. of the equation :

$$\begin{aligned}\Rightarrow \frac{\text{cos}^2 A}{\text{1 - tan A}} + \frac{\text{sin}^3 A}{\text{sin A - cos A}} \\[1em] \Rightarrow \frac{\text{cos}^2 A}{1 - \frac{\text{sin A}}{\text{cos A}}} + \frac{\text{sin}^3 A}{\text{sin A - cos A}} \\[1em] \Rightarrow \frac{\text{cos}^2 A}{\frac{\text{cos A - sin A}}{\text{cos A}}} + \frac{\text{sin}^3 A}{\text{sin A - cos A}} \\[1em] \Rightarrow \frac{\text{cos}^3 A}{\text{cos A - sin A}} + \frac{\text{sin}^3 A}{\text{sin A - cos A}} \\[1em] \Rightarrow -\frac{\text{cos}^3 A}{\text{sin A - cos A}} + \frac{\text{sin}^3 A}{\text{sin A - cos A}} \\[1em] \Rightarrow \frac{\text{sin}^3 A - \text{cos}^3 A}{\text{sin A - cos A}}\end{aligned}$$

By formula,

a³ - b³ = (a - b)(a² + ab + b²)

$$\begin{aligned}\Rightarrow \frac{\text{(sin A - cos A)(sin}^2 A + \text{sin A cos A} + \text{cos}^2 A)}{\text{sin A - cos A}} \\[1em] \Rightarrow \text{sin}^2 A + \text{cos}^2 A + \text{sin A cos A}\end{aligned}$$

By formula,

sin² A + cos² A = 1.

⇒ 1 + sin A cos A.

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{cos}^2 A}{\text{1 - tan A}} + \frac{\text{sin}^3 A}{\text{sin A - cos A}}$ = 1 + sin A cos A.

(v) Solving L.H.S. of the equation :

$\Rightarrow \frac{\text{1 + sin A - cos A}}{\text{sin A + cos A - 1}}$

Multiplying numerator and denominator by (1 + sin A + cos A) :

$$\begin{aligned}\Rightarrow \frac{\text{(1 + sin A - cos A)(1 + sin A + cos A)}}{\text{(sin A + cos A - 1)(sin A + cos A + 1)}} \\[1em] \Rightarrow \frac{\text{(1 + sin A)}^2 - \text{cos}^2 A}{\text{(sin A + cos A)}^2 - 1} \\[1em] \Rightarrow \frac{1 + \text{2 sin A} + \text{sin}^2 A - \text{cos}^2 A}{\text{sin}^2 A + \text{cos}^2 A + \text{2 sin A cos A} - 1}\end{aligned}$$

By formula,

cos² A = 1 - sin² A and sin² A + cos² A = 1.

$$\begin{aligned}\Rightarrow \frac{1 + \text{2 sin A} + \text{sin}^2 A - (1 - \text{sin}^2 A)}{1 + \text{2 sin A cos A} - 1} \\[1em] \Rightarrow \frac{\text{2 sin A} + \text{2 sin}^2 A}{\text{2 sin A cos A}} \\[1em] \Rightarrow \frac{\text{2 sin A(1 + sin A)}}{\text{2 sin A cos A}} \\[1em] \Rightarrow \frac{\text{1 + sin A}}{\text{cos A}}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{sin A + 1 - cos A}}{\text{cos A - 1 + sin A}} = \frac{\text{1 + sin A}}{\text{cos A}}$.

Given,

⇒ sin A + cos A = $\sqrt{3}$

Squaring both sides :

⇒ (sin A + cos A)² = $(\sqrt{3})^2$

⇒ sin² A + cos² A + 2 sin A cos A = 3

By formula,

sin² A + cos² A = 1.

⇒ 1 + 2 sin A cos A = 3

⇒ 2 sin A cos A = 2

⇒ sin A cos A = 1.

Now, solving tan A + cot A :

$$\begin{aligned}\Rightarrow \text{tan A + cot A} \\[1em] \Rightarrow \frac{\text{sin A}}{\text{cos A}} + \frac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \frac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}} \\[1em] \Rightarrow \frac{1}{\text{sin A cos A}} \\[1em] \Rightarrow \frac{1}{1} \\[1em] \Rightarrow 1.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that tan A + cot A = 1.

(i) By formula,

sin (90° - A) = cos A and cos (90° - A) = sin A.

Substituting above values in L.H.S. :

$$\begin{aligned}\Rightarrow \frac{\text{sin A}}{\text{cos A}} + \frac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \frac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}}\end{aligned}$$

By formula,

sin² A + cos² A = 1.

$$\begin{aligned}\Rightarrow \frac{1}{\text{sin A cos A}} \\[1em] \Rightarrow \frac{1}{\text{sin A}} \times \frac{1}{\text{cos A}} \\[1em] \Rightarrow \text{cosec A sec A}.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{sin A}}{\text{sin (90° - A)}} + \frac{\text{cos A}}{\text{cos (90° - A)}}$ = sec A cosec A.

(ii) Solving L.H.S. of the equation :

$\Rightarrow \text{2 cos}^2 A + \frac{2}{\text{1 + cot}^2 A}$

By formula,

1 + cot² A = cosec² A.

$$\begin{aligned}\Rightarrow \text{2 cos}^2 A + \frac{2}{\text{cosec}^2 A} \\[1em] \Rightarrow \text{2 cos}^2 A + \text{2 sin}^2 A \\[1em] \Rightarrow 2(\text{cos}^2 A + \text{sin}^2 A)\end{aligned}$$

By formula,

sin² A + cos² A = 1.

⇒ 2 × 1

⇒ 2.

Since, L.H.S. = R.H.S.

Hence, proved that $\text{2 cos}^2 A + \frac{2}{\text{1 + cot}^2 A}$ = 2.

(i) Solving,

⇒ sec² A(1 + sin A)(1 - sin A)

⇒ sec² A(1 - sin² A)

By formula,

1 - sin² A = cos² A.

$$\begin{aligned}\Rightarrow \text{sec}^2 A \text{ cos}^2 A \\[1em] \Rightarrow \frac{1}{\text{cos}^2 A} \times \text{cos}^2 A \\[1em] \Rightarrow 1.\end{aligned}$$

Hence, sec² A(1 + sin A)(1 - sin A) = 1.

(ii) Given,

cosec A = 2x and tan A = $\frac{x}{2}$

⇒ cot A = $\frac{1}{\text{tan A}} = \frac{2}{x}$

By formula,

cosec² A - cot² A = 1

$$\begin{aligned}\Rightarrow (2x)^2 - \Big(\frac{2}{x}\Big)^2 = 1 \\[1em] \Rightarrow 4x^2 - \frac{4}{x^2} = 1 \\[1em] \Rightarrow 4\Big(x^2 - \frac{1}{x^2}\Big) = 1 \\[1em] \Rightarrow x^2 - \frac{1}{x^2} = \frac{1}{4}.\end{aligned}$$

Hence, $x^2 - \frac{1}{x^2} = \frac{1}{4}$.

(i) Solving, for x = 60° :

$$\begin{aligned}\Rightarrow \text{tan x + cot x} \\[1em] \Rightarrow \text{tan 60° + cot 60°} \\[1em] \Rightarrow \sqrt{3} + \frac{1}{\sqrt{3}} \\[1em] \Rightarrow \frac{3 + 1}{\sqrt{3}} \\[1em] \Rightarrow \frac{4}{\sqrt{3}} \\[1em] \Rightarrow \frac{4 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\[1em] \Rightarrow \frac{4\sqrt{3}}{3}.\end{aligned}$$

Hence, A = tan x + cot x = $\frac{4\sqrt{3}}{3}$.

(ii) Solving, for x = 60° :

$$\begin{aligned}\Rightarrow \frac{1}{\text{sin x cos x}} \\[1em] \Rightarrow \frac{1}{\text{sin 60° cos 60°}} \\[1em] \Rightarrow \frac{1}{\frac{\sqrt{3}}{2} \times \frac{1}{2}} \\[1em] \Rightarrow \frac{1}{\frac{\sqrt{3}}{4}} \\[1em] \Rightarrow \frac{4}{\sqrt{3}} \\[1em] \Rightarrow \frac{4 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\[1em] \Rightarrow \frac{4\sqrt{3}}{3}.\end{aligned}$$

Hence, B = $\frac{1}{\text{sin x cos x}} = \frac{4\sqrt{3}}{3}$.

(iii) From parts (i) and (ii), A = $\frac{4\sqrt{3}}{3}$ and B = $\frac{4\sqrt{3}}{3}$.

∴ A = B.

Hence, the relation between A and B is A = B.

(iv) Since A = B, that is tan x + cot x = $\frac{1}{\text{sin x cos x}}$ for all values of x in the domain.

Hence, the corresponding identity is tan x + cot x = $\frac{1}{\text{sin x cos x}}$.