Trigonometrical Identities

Solving L.H.S:

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

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

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

Solving L.H.S:

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

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

Hence, proved that $\Big(\frac{\sec A - 1}{\sec A + 1}\Big) = \Big(\frac{1 - \cos A}{1 + \cos A}\Big)$.

Solving L.H.S of the equation:

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

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

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

Solving L.H.S. of the equation :

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

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

Hence, proved that $\Big(\frac{1}{\tan A + \cot A}\Big) = \cos A \times \sin A$.

Solving L.H.S. of the equation :

⇒ (1 + tan A)² + (1 - tan A)²

⇒ 1 + tan²A + 2 tan A + 1 + tan² A - 2 tan A

⇒ 2 + 2 tan²A

⇒ 2(1 + tan²A)

By formula,

1 + tan²A = sec² A

⇒ 2sec² A

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

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

Solving L.H.S:

⇒ (sin² θ - 1) (tan² θ + 1) + 1

By formula,

⇒ sin²θ − 1 = − cos²θ

⇒ tan²θ+ 1 = sec²θ

= -cos²θ (sec²θ) + 1

By formula,

⇒ cos²θ × sec²θ = 1

= −1 + 1

= 0

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

Hence, proved that (sin² θ - 1) (tan² θ + 1) + 1 = 0.

Solving L.H.S. of the equation :

⇒ cosec A(1 + cos A)(cosec A - cot A)

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

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

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

Solving L.H.S. of the equation :

⇒ sec A(1 - sin A)(sec A + tan A)

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

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 :

⇒ (cosec θ - sin θ)(sec θ - cos θ)(tan θ + cot θ) = 1

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

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

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

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.

By formula,

sec² A = 1 + tan² A

cos² A = 1 - sin² A

Solving L.H.S. of the equation

⇒ (sec A - cos A)(sec A + cos A)

⇒ sec² A - cos² A

⇒ 1 + tan² A - (1 - sin² A)

⇒ 1 - 1 + tan² A + sin² A

⇒ sin² A + tan² A.

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

Hence, proved that (sec A + cos A)(sec A - cos A) = sin² A + tan² A.

Solving L.H.S of equation,

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

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

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

Solving L.H.S of equation,

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

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

Hence, proved that cot² A - cos² A = cos² A cot² A.

Solving L.H.S of equation,

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

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

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

Solving L.H.S of equation,

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

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

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

Solving L.H.S of equation,

⇒ sin A (1 + tan A) + cos A (1 + cot A)

⇒ sin A + sin A tan A + cos A + cos A cotA

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

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

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

Solving L.H.S of equation,

$\Big(\frac{1}{1 + \tan^2 A}\Big) + \Big(\frac{1}{1 + \cot^2 A}\Big) = 1$

By formula:

1 + tan² A = sec² A

1 + cot² A = cosec² A

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

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

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

Solving L.H.S. of the equation :

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

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

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

Solving L.H.S,

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

Since,

L.H.S = R.H.S

Hence, proved that

$\frac{\sin A - \sin^3 A}{\cos^3 A - \cos A} \times (\sec A - \cosec A) = \cosec A(\cot A - 1)$.

Solving L.H.S. of the equation :

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

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

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

Solving L.H.S. of the equation :

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

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

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

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 × $\frac{1}{\sin A}$ + cos² A + 1 + tan² A + 2 × cos A × $\frac{1}{\cos A}$

⇒ 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.

Factorizing,

⇒ sin³ θ + cos³ θ = (sin θ + cos θ)(sin² θ + cos² θ - sin θ cos θ)

⇒ sin³ θ + cos³ θ = (sin θ + cos θ)(1 - sin θ cos θ) ...........(1)

To prove,

$\frac{\text{sin}^3 θ + \text{cos}^3 θ}{\text{sin θ + cos θ}}$ + sin θ cos θ = 1

Substituting value of sin³ θ + cos³ θ from equation (1) in L.H.S. of above equation :

⇒ $\frac{\text{(sin θ + cos θ)(1 - sin θ cos θ)}}{\text{sin θ + cos θ}} + \text{sin θ cos θ}$

⇒ 1 - sin θ cos θ + sin θ cos θ

⇒ 1.

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

Hence, proved that $\frac{\text{sin}^3 θ + \text{cos}^3 θ}{\text{sin θ + cos θ}} + \text{sin θ cos θ} = 1$.

Solving L.H.S of equation,

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

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

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

The L.H.S of above equation can be written as,

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

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

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

Solving L.H.S of the equation,

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

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

Hence, proved that $\Big(\frac{\tan \theta + \sin \theta}{\tan \theta - \sin \theta}\Big) = \Big(\frac{\sec \theta + 1}{\sec \theta - 1}\Big)$.

L.H.S. of the equation can be written as,

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

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

Hence, proved that $\Big(\frac{\cot \theta + \cosec \theta - 1}{\cot \theta - \cosec \theta + 1}\Big) = \Big(\frac{1 + \cos \theta}{\sin \theta}\Big)$.

L.H.S. of the equation can be written as,

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

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

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

The equation can be written as,

$\frac{1}{\sec A + \tan A} + \frac{1}{\sec A - \tan A} = \frac{2}{\cos A}$

L.H.S. of the equation can be written as,

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

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

Hence, proved that

$\Big(\frac{1}{\sec A + \tan A}\Big) - \Big(\frac{1}{\cos A}\Big) = \Big(\frac{1}{\cos A}\Big) - \Big(\frac{1}{\sec A - \tan A}\Big)$.

L.H.S. of the equation can be written as,

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

Multiplying numerator and denominator by (cos A - 1), we get :

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

R.H.S. of the equation can be written as,

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

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

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

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

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

L.H.S. of the equation can be written as,

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

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

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

The L.H.S. of the equation can be written as,

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

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

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

The L.H.S. of the equation can be written as,

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

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

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

The L.H.S. of the equation can be written as,

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

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

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

The L.H.S. of the equation can be written as,

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

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

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

The L.H.S of above equation can be written as,

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

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

Hence, proved that $\Big(\frac{1 - \tan \theta}{1 - \cot \theta}\Big)^2 = \tan^2 \theta$.

The L.H.S of above equation can be written as,

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

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

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

Solving L.H.S.,

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

Hence, $\frac{(\cot A + \tan A - 1)(\sin A + \cos A)}{\sin^{3}A + \cos^{3}A} = \sec A \times \cosec A$.

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

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

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

Hence, proved that $(\sin \theta + \cos \theta)(\tan \theta + \cot \theta) = \sec \theta + \cosec \theta$.

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

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

Hence, proved that (sin θ + cos θ)(cosec θ - sec θ) = cosec θ.sec θ - 2 tan θ.

Solving L.H.S. of the equation :

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

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

Hence, proved that $(1 + \tan^2 A) + (1 + \cot^2 A) = \Big(\frac{1}{\sin^2 A - \sin^4 A}\Big)$.

Solving L.H.S. of the equation :

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

Solving R.H.S. of the equation :

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

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

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

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

⇒ (tan A + cot A)(cosec A - sin A)(sec A - cos A)

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

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

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

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

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

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

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

Given,

x = a cosec θ, y = b cot θ

⇒ cosec θ = $\frac{x}{a}$

⇒ cot θ = $\frac{y}{b}$

Using the identity

cosec² θ - cot² θ = 1

Substitute the values:

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

Hence, the required relation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

x = a cot θ + b cosec θ .....(1)

y = a cosec θ + b cot θ .....(2)

Subtract equation (2) from (1):

⇒ x - y = a cot θ + b cosec θ - (a cosec θ + b cot θ)

⇒ x - y = a cot θ + b cosec θ - a cosec θ - b cot θ

⇒ x - y = a (cot θ - cosec θ) - b (cot θ - cosec θ)

⇒ x - y = (a - b) (cot θ - cosec θ)

Add equation (2) from (1):

⇒ x + y = a cot θ + b cosec θ + (a cosec θ + b cot θ)

⇒ x + y = a (cot θ + cosec θ) + b (cot θ + cosec θ)

⇒ x + y = (a + b) (cot θ + cosec θ)

Now multiply:

⇒ (x + y)(x - y) = (a + b)(a - b)(cot θ + cosec θ)(cot θ - cosec θ)

⇒ x² - y² = (a² - b²)(cot² θ - cosec² θ)

By identity: cot² θ − cosec ² θ = −1

⇒ x² - y² = - (a² - b²)

⇒ x² - y² = b² - a²

Hence, the required relation is x² - y² = b² - a² .

Given,

⇒ x = a sec³ θ

⇒ sec³ θ = $\frac{x}{a}$

⇒ sec² θ = $\Big(\frac{x}{a}\Big)^\frac{2}{3}$

⇒ y = b tan³ θ

⇒ tan³ θ = $\frac{y}{b}$

⇒ tan² θ = $\Big(\frac{y}{b}\Big)^\frac{2}{3}$

Using the identity

sec² θ - tan² θ = 1

Substitute,

$\Big(\frac{x}{a}\Big)^\frac{2}{3} - \Big(\frac{y}{b}\Big)^\frac{2}{3} = 1$

Hence,the required relation is $\Big(\frac{x}{a}\Big)^\frac{2}{3} - \Big(\frac{y}{b}\Big)^\frac{2}{3} = 1$.

$$\begin{aligned}\Big(\frac{x}{a}\Big) \cos \theta + \Big(\frac{y}{b}\Big) \sin \theta = 1.....(1)\\[1em] \Big(\frac{x}{a}\Big) \sin \theta - \Big(\frac{y}{b}\Big) \cos \theta = -1.....(2)\end{aligned}$$

Square and add equations (1) and (2):

$$\begin{aligned}\Rightarrow \Big[\Big(\frac{x}{a}\Big) \cos \theta + \Big(\frac{y}{b}\Big) \sin \theta\Big]^2 + \Big[\Big(\frac{x}{a}\Big) \sin \theta - \Big(\frac{y}{b}\Big) \cos \theta\Big]^2 = 1^2 + (-1)^2 \\[1em] \Rightarrow \Big[\Big(\frac{x}{a}\Big)^2 \cos^2 \theta + \Big(\frac{y}{b}\Big)^2 \sin^2 \theta + 2\frac{xy}{ab} \sin \theta \cos \theta \Big] + \Big[\Big(\frac{x}{a}\Big)^2 \sin^2 \theta + \Big(\frac{y}{b}\Big)^2 \cos^2 \theta - 2\frac{xy}{ab} \sin \theta \cos \theta\Big] = 2 \\[1em] \Rightarrow \Big(\frac{x}{a}\Big)^2 \cos^2 \theta + \Big(\frac{y}{b}\Big)^2 \sin^2 \theta + \Big(\frac{x}{a}\Big)^2 \sin^2 \theta + \Big(\frac{y}{b}\Big)^2 \cos^2 \theta = 2 \\[1em] \Rightarrow \Big(\frac{x}{a}\Big)^2 (\cos^2 \theta + \sin^2 \theta) + \Big(\frac{y}{b}\Big)^2 (\sin^2 \theta + \cos^2 \theta) = 2 \\[1em] \Rightarrow \Big(\frac{x}{a}\Big)^2 + \Big(\frac{y}{b}\Big)^2 = 2 .\end{aligned}$$

Hence, the required relation is $\Big(\frac{x}{a}\Big)^2 + \Big(\frac{y}{b}\Big)^2 = 2$.

⇒ x = h + a cos θ

⇒ cos θ = $\frac{x - h}{a}$.....(1)

⇒ y = k + b sin θ

⇒ sin θ = $\frac{y - k}{b}$.....(2)

Square and add equations (1) and (2):

$$\begin{aligned}\Rightarrow \Big(\frac{x - h}{a}\Big)^2 + \Big(\frac{y - k}{b}\Big)^2 = \cos^2 \theta + \sin^2 \theta \\[1em] \Rightarrow \Big(\frac{x - h}{a}\Big)^2 + \Big(\frac{y - k}{b}\Big)^2 = 1 \\[1em] \Rightarrow \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 .\end{aligned}$$

Hence,the required relation is $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$.

To prove:

(m² + n²)cos²B = n²

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

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

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

Hence, proved that (m² + n²)cos²B = n².

⇒ x = a sec A cos B

sec A cos B = $\frac{x}{a}$....(1)

⇒ y = b sec A sin B

sec A sin B = $\frac{y}{b}$....(2)

Square and add equations (1) and (2) :

$$\begin{aligned}\Rightarrow \frac{x^2}{a^2} + \frac{y^2}{b^2} = \sec^2 A \cos^2 B + \sec^2 A \sin^2 B \\[1em] \Rightarrow \frac{x^2}{a^2} + \frac{y^2}{b^2} = \sec^2 A (\cos^2 B + \sin^2 B) \\[1em] \Rightarrow \frac{x^2}{a^2} + \frac{y^2}{b^2} = \sec^2 A ...(3)\end{aligned}$$

Now,

⇒ z = c tan A

tan A = $\frac{z}{c}$

tan² A = $\frac{z^2}{c^2}$...(4)

Subtract (4) from (3):

$$\begin{aligned}\Rightarrow \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = \sec^2 A - \tan^2 A \\[1em] \Rightarrow \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1.\end{aligned}$$

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

(i) Given,

sin 83° 12′

⇒ sin 83° 12′

From table we have,

∴ sin 83° 12′ = 0.9930

Mean difference of 0′ = 0.0000

Hence, sin 83° 12′ = 0.9930.

(ii) Given,

sin 61° 12′

⇒ sin 61° 12′ + 1'

From table we have,

sin 61° 12′ = 0.8763

Mean difference of 2' = .0003 (To be added)

sin 61° 14′ = 0.8763 + 0.0003 = 0.8766

Hence, sin 61° 14′ = 0.8766.

(iii) Given,

sin 9° 55′

⇒ sin 9° 54′ + 1'

From table we have,

sin 9° 54′ = 0.1719

Mean difference of 1' = .0003 (To be added)

sin 9° 55′ = 0.1719 + 0.0003 = 0.1722

Hence, sin 9° 55′ = 0.1722.

(iv) Given,

sin 32° 40′

⇒ sin 32° 36′ + 4'

From table we have,

sin 32° 36′ = 0.5388

Mean difference of 4' = .0010 (To be added)

sin 32° 40′ = 0.5388 + 0.0010 = 0.5398

Hence, sin 32° 40′ = 0.5398.

(i) Given,

cos 48° 36′

⇒ cos 48° 36′ = cos 48° 36′

From table we have,

cos 48° 36′ = 0.6613

Mean difference of 0′ = 0.0000

Therefore,

cos 48° 36′ = 0.6613

Hence, cos 48° 36′ = 0.6613.

(ii) Given,

cos 23° 6′

⇒ cos 23° 6′ = cos 23° 6′

From table we have,

cos 23° 6′ = 0.9198

Mean difference of 0′ = 0.0000

Therefore,

cos 23° 6′ = 0.9198

Hence, cos 23° 6′ = 0.9198.

(iii) Given,

cos 70° 17′

⇒ cos 70° 18′ − 1′

From table we have,

cos 70° 18′ = 0.3371

Mean difference of 1′ = 0.0003 (to be added)

Therefore,

cos 70° 17′ = 0.3371 + 0.0003

cos 70° 17′ = 0.3374

Hence, cos 70° 17′ = 0.3374.

(iv) Given,

cos 85° 8′

From table we have,

cos 85° 6′ = 0.0854

Mean difference of 2′ = 0.0006 (to be subtracted)

Therefore,

cos 85° 8′ = 0.0854 − 0.0006

cos 85° 8′ = 0.0848

Hence, cos 85° 8′ = 0.0848.

(i) Given,

tan 24° 24′

⇒ tan 24° 24′

From table we have,

tan 24° 24′ = 0.4536

Mean difference of 0′ = 0.0000

Therefore,

tan 24° 24′ = 0.4536

Hence, tan 24° 24′ = 0.4536.

(ii) Given,

tan 9° 38′

⇒ tan 9° 36′ + 2'

From table we have,

tan 9° 36′ = 0.1691

Mean difference of 2′ = 0.0006 (to be added)

Therefore,

tan 9° 38′ = 0.1691 + 0.0006 = 0.1697

Hence, tan 9° 38′ = 0.1697.

(iii) Given,

tan 31° 27′

⇒ tan 31° 24′ + 3'

From table we have,

tan 31° 24′ = 0.6104

Mean difference of 3′ = 0.0012 (to be added)

Therefore,

tan 31° 27′ = 0.6104 + 0.0012 = 0.6116

Hence, tan 31° 27′ = 0.6116.

(iv) Given,

tan 65° 50′

⇒ tan 65° 48′ + 2'

From table we have,

tan 65° 48′ = 2.2251

Mean difference of 2′ = 0.0034 (to be added)

Therefore,

tan 65° 50′ = 2.2251 + 0.0034 = 2.2285

Hence, tan 65° 50′ = 2.2285.

(i) Given,

sin θ = 0.36

sin 21° 6' = .3600 (From tables)

Difference = .0000

Mean difference for 0' = .0000

Hence, θ = 21° 6'.

(ii) Given,

sin θ = 0.4274

sin 25° 18' = 0.4274 (From tables)

Difference = .0000

Mean difference for 0' = .0000

Hence, θ = 25° 18'.

(iii) Given,

sin θ = 0.5955

sin 36° 30' = 0.5948 (From tables)

Difference = .0007

Mean difference for 3' = .0007

θ = 36° 30'+ 3' = 36° 33'

Hence, θ = 36° 33'.

(iv) Given,

sin θ = 0.8229

sin 55° 18' = 0.8221 (From tables)

Difference = .0008

Mean difference for 5' = .0008

θ = 55° 18'+ 5' = 55° 23'

Hence, θ = 55° 23'.

(i) Given,

sin θ = 0.42

sin 24° 48'= 0.4195 (From tables)

Difference = .0005

Mean difference for 2' = .0005

θ = 24° 48'+ 2' = 24° 50'

Hence, θ = 24° 50'.

(ii) cos θ

cos 24° 48' = 0.9078

Mean difference for 2' = .0002 (to be subtracted)

Therefore,

cos 24° 50' = 0.9078 - 0.0002 = 0.9076

Hence, cos 24° 50' = 0.9076.

(iii) tan θ

tan 24° 48' = 0.4621

Mean difference for 2' = .0007 (to be added)

Therefore,

tan 24° 48' = 0.4621 + 0.0007 = 0.4628

Hence, tan 24° 50' = 0.4628.

(i) Given,

cos θ = 0.94

cos 19° 54' = .9403

Difference = .0003

Mean difference of 3' = .0003

θ = 19° 54' + 3' = 19° 57'.

Hence, θ = 19° 57'.

(ii) Given,

cos θ = 0.8092

cos 35° 54' = .8100

Difference = .0008

Mean difference of 5' = .0008

θ = 35° 54' + 5' = 35° 59'.

Hence, θ = 35° 59'.

(iii) Given,

cos θ = 0.1679

cos 80° 18' = 0.1685

Difference = .0006

Mean difference of 2' = .0006

θ = 80° 18' + 2' = 80° 20'.

Hence, θ = 80° 20'.

(i) Given,

cos θ = 0.51

cos 59° 18' = 0.5105

Difference = 0.5105 − 0.5100 = 0.0005

Mean difference of 2' = 0.0005

θ = 59° 18' + 2' = 59° 20'.

Hence, θ = 59° 20'.

(ii) Given,

sin θ

sin 59° 18' = 0.8599

Mean difference of 2' = 0.0003

sin 59° 20' = 0.8599 + 0.0003 = 0.8602.

Hence, sin θ = 0.8602.

(iii) Given,

tan θ

tan 59° 18' = 1.6842

Mean difference of 2' = 0.0022

tan 59° 20' = 1.6842 + 0.0022 = 1.6864.

Hence, tan θ = 1.6864.

(i) Given,

tan θ = 1.476

tan 55° 54' = 1.4770

Difference = .0010

Mean difference of 1' = .0009(to be subtracted)

θ = 55° 54' - 1' = 55° 53'.

Hence, θ = 55° 53'.

(ii) Given,

tan θ = 2.91

tan 71° = 2.9042

Difference = .0058

Mean difference of 2' = .0058(to be added)

θ = 71° + 2' = 71° 2'.

Hence, θ = 71° 2'.

(iii) Given,

tan θ = 0.3

tan 16° 42' = 0.3

Difference = 0.000

Mean difference of 0' = .000

θ = 16° 42'.

Hence, θ = 16° 42'.

tan A = $\frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AB} = \frac{8}{15}$

Hence, option 1 is the correct option.

Given,

sin θ = $\Big(\frac{5}{13}\Big) = \frac{\text{opposite}}{\text{hypotenuse}}$

Opposite = 5, Hypotenuse = 13

tan A = $\frac{\text{opposite}}{\text{adjacent}} = \frac{5}{12}$

Hence, option 1 is the correct option.

sec θ = $\Big(\frac{25}{7}\Big) = \frac{\text{hypotenuse}}{\text{base}}$

Perpendicular = $\sqrt{25^2 - 7^2} = \sqrt{576}$ = 24.

cot θ = $\frac{\text{Base}}{\text{perpendicular}} = \Big(\frac{7}{24}\Big)$.

Hence, option 3 is the correct option.

Given,

⇒ (1 + tan²θ)(1 − sin θ)(1 + sin θ)

⇒ sec²θ (1 - sin²θ)

⇒ sec²θ cos²θ

⇒ 1

Hence, option 2 is the correct option.

Let ABC be a right angle triangle with ∠B = 90° and ∠C = θ.

By formula,

$\sin \theta = \frac{\text{perpendicular}}{\text{hypotenuse}}$

Substituting values we get :

$\frac{a}{b} = \frac{AB}{AC}$

Let AB = ak and AC = bk.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ (bk)² = (ak)² + BC²

⇒ b²k² = a²k² + BC²

⇒ BC² = b²k² - a²k²

⇒ BC = k$\sqrt{b^2 - a^2}$

By formula,

$$\begin{aligned}\cos \theta = \frac{\text{base}}{\text{hypotenuse}} \\[1em] = \frac{BC}{AC} \\[1em] = \frac{k\sqrt{b^2 - a^2}}{bk} \\[1em] = \frac{\sqrt{b^2 - a^2}}{b} .\end{aligned}$$

Hence, option 4 is the correct option.

We know that,

$$\begin{aligned}cot θ = \frac{\text{base}}{\text{perpendicular}} \\[1em] \Rightarrow \frac{\cot y}{\cot x} = \frac{\frac{AC}{BC}}{\frac{AC}{CD}} \\[1em] \Rightarrow \frac{\cot y}{\cot x} = \frac{CD}{BC} \\[1em] \Rightarrow \frac{\cot y}{\cot x} = \frac{CD}{2CD} \\[1em] \Rightarrow \frac{\cot y}{\cot x} = \frac{1}{2} .\end{aligned}$$

Hence, option 1 is the correct option.

Given,

(sin A + cos A) sec A

$$\begin{aligned}\Rightarrow (\sin A + \cos A)\frac{1}{\cos A} \\[1em] \Rightarrow \frac{\sin A}{\cos A} + \frac{\cos A}{\cos A} \\[1em] \Rightarrow \tan A + 1 \\[1em] \Rightarrow \frac{5}{12} + 1 \\[1em] \Rightarrow \frac{5 + 12}{12} \\[1em] \Rightarrow \frac{17}{12}.\end{aligned}$$

Hence, option 3 is the correct option.

Given,

3 cos θ = 1

We know that,

⇒ sin² θ = 1 - cos² θ

$$\begin{aligned}\Rightarrow \sin^2 \theta = 1 - \frac{1}{9} \\[1em] \Rightarrow \sin^2 \theta = \frac{9 - 1}{9} \\[1em] \Rightarrow \sin^2 \theta = \frac{8}{9} \\[1em] \Rightarrow \sin \theta = \frac{2\sqrt{2}}{3} \\[1em] \Rightarrow \cosec \theta = \frac{1}{\sin \theta} = \frac{3}{2\sqrt{2}}.\end{aligned}$$

Hence, option 2 is the correct option.

Given,

x cos A = 1

cos A = $\frac{1}{x}$

⇒ sec A = x

tan A = y

We know that

sec² A - tan² A = 1

∴ x² − y² = 1

Hence, option 2 is the correct option.

ST = PS − RQ = 14 − 5 = 9 cm

In ΔSTR,

$TR = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12.$

We know that,

$$\begin{aligned}\tan \alpha = \frac{\text{opposite}}{\text{adjacent}} \\[1em] \tan \alpha = \frac{TR}{ST} \\[1em] \tan \alpha = \frac{12}{9} \\[1em] \tan \alpha = \frac{4}{3}.\end{aligned}$$

Hence, option 1 is the correct option.

Given,

(1 + tan θ + sec θ)(1 + cot θ − cosec θ)

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

Hence, option 4 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{1}{5} \cot^2\theta + \frac{1}{5}\Big) \\[1em] \Rightarrow \frac{1}{5} \Big(\cot^2\theta + 1 \Big) \\[1em] \Rightarrow \frac{1}{5} (\cosec^2 \theta).\end{aligned}$$

We know that,

$$\begin{aligned}\cosec \theta = \frac{1}{\sin \theta } \\[1em] \Rightarrow \cosec \theta = \frac{1}{\frac{1}{2}} = 2\\[1em] \Rightarrow \cosec^2 \theta = 2^2 = 4 \\[1em] \Rightarrow \frac{1}{5}\cosec^2 \theta \\[1em] \Rightarrow \frac{1}{5}(4) \\[1em] \Rightarrow \frac{4}{5}.\end{aligned}$$

Hence, option 2 is the correct option.

cos θ = $\frac{2}{3}$

sec θ = $\frac{3}{2}$

sec² θ = $\frac{9}{4}$

tan² θ = sec² θ - 1 = $\frac{9}{4} - 1 = \frac{5}{4}$

Given,

⇒ 2 sec²θ + 2 tan²θ − 7

$$\begin{aligned}\Rightarrow 2\Big(\frac{9}{4}\Big) + 2\Big(\frac{5}{4}\Big) - 7 \\[1em] \Rightarrow \Big(\frac{18}{4}\Big) + \Big(\frac{10}{4}\Big) - 7 \\[1em] \Rightarrow \Big(\frac{28}{4}\Big) - 7 \\[1em] \Rightarrow 7 - 7 \\[1em] \Rightarrow 0.\end{aligned}$$

Hence, option 1 is the correct option.

Given,

24 cot θ = 7

cot θ = $\frac{7}{24} = \frac{\text{adjacent}}{\text{opposite}}$

We know that,

$$\begin{aligned}\Rightarrow \text{Hypotenuse} = \sqrt{\text{opposite}^2 + \text{adjacent}^2 } \\[1em] = \sqrt{(24)^2 + (7)^2} \\[1em] = \sqrt{576 + 49} \\[1em] = \sqrt{625} \\[1em] = 25.\end{aligned}$$

Now,

sin θ = $\frac{\text{opposite}}{\text{hypotenuse}} = \frac{24}{25}$

Hence, option 2 is the correct option.

Given,

tan θ + cot θ = 2

Square on both sides,

⇒ (tan θ + cot θ)² = 2²

⇒ tan² θ + cot² θ + 2 tan θ cot θ = 4

⇒ tan² θ + cot² θ + 2 tan θ $\frac{1}{\tan \theta}$ = 4

⇒ tan² θ + cot² θ + 2 = 4

⇒ tan² θ + cot² θ = 4 - 2

⇒ tan² θ + cot² θ = 2

Hence, option 2 is the correct option.

Given,

cot A + $\Big(\frac{1}{\cot A}\Big)$ = 2

Square on both sides,

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

Hence, option 3 is the correct option.

tan θ = $\frac{3}{4}$

Then,

sin θ = $\frac{3}{5}$, cos θ = $\frac{4}{5}$

Substitute,

$$\begin{aligned}\Rightarrow \frac{4\Big(\frac{3}{5}\Big) - 3\Big(\frac{4}{5}\Big)}{4\Big(\frac{3}{5}\Big) + 3\Big(\frac{4}{5}\Big)} \\[1em] \Rightarrow \frac{\Big(\frac{12}{5}\Big) - \Big(\frac{12}{5}\Big)}{\Big(\frac{12}{5}\Big) + \Big(\frac{12}{5}\Big)} \\[1em] \Rightarrow \frac{0}{\frac{24}{5}} \\[1em] \Rightarrow 0.\end{aligned}$$

Hence, option 1 is the correct option.

Given,

tan θ = $\Big(\frac{1}{\sqrt{7}}\Big)$

tan² θ = $\Big(\frac{1}{7}\Big)$

We know that,

$$\begin{aligned}\sec^2 \theta = 1 + \tan^2 \theta \\[1em] \sec^2 \theta = 1 + \frac{1}{7} \\[1em] \sec^2 \theta = \frac{8}{7}. \\[1em] \cosec^2 \theta = 1 + \cot^2 \theta \\[1em] \cosec^2 \theta = 1 + 7 \\[1em] \cosec^2 \theta = 8.\end{aligned}$$

Given expression,

$$\begin{aligned}\Rightarrow \Big(\frac{\cosec^2\theta + \sec^2\theta}{\cosec^2\theta - \sec^2\theta}\Big) \\[1em] \Rightarrow \frac{8 + \frac{8}{7}}{8 - \frac{8}{7}} \\[1em] \Rightarrow \frac{\frac{56 + 8}{7}}{\frac{56 - 8}{7}} \\[1em] \Rightarrow \frac{64}{48} = \frac{4}{3}.\end{aligned}$$

Hence, option 2 is the correct option.

⇒ (1 + sin A)(1 − sin A)

⇒ 1 − sin² A

⇒ cos²A

Hence, option 4 is the correct option.

1 + cot² A = cosec² A

$\sqrt{1 + \cot^2 A}$ = cosec A

sin A = $\frac{1}{\cosec A}$

sin A = $\frac{1}{\sqrt{1 + \cot^2 A}}$

Hence, option 1 is the correct option.

sec θ = 2x

x = $\frac{\sec \theta}{2}$

y tan θ = 2

$\frac{1}{y} = \frac{\tan \theta}{2}$

We have,

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

Hence, option 1 is the correct option.

    a - b = 3 sec² θ - (3 tan² θ - 2)

= 3 sec² θ - 3 tan² θ + 2

= 3(sec² θ - tan² θ) + 2

= 3(1) + 2

= 3 + 2

= 5.

Hence, option 4 is the correct option.

⇒ (cos⁴ θ − sin⁴ θ)

⇒ (cos² θ − sin² θ)(cos² θ + sin² θ)

⇒ (cos² θ - sin² θ)

⇒ (cos² θ - 1 + cos² θ)

⇒ (2cos² θ - 1)

Hence, option 2 is the correct option.

We know that,

cosec² θ − cot² θ = 1

cosec θ − cot θ (cosec θ + cot θ)= 1

$\frac{1}{3}$ (cosec θ + cot θ)= 1

(cosec θ + cot θ)= 3

Hence, option 3 is the correct option.

sin θ = cos θ

Divide both sides by cos θ

$\frac{\sin \theta}{\cos \theta} = \frac{\cos \theta}{\cos \theta}$

tan θ = 1

For acute angles, tan θ = 1 when θ = 45°.

⇒ sin 45° + cos 45°

$\frac{1}{\sqrt2} + \frac{1}{\sqrt2}$

⇒ $\frac{2}{\sqrt2}$

⇒ $\sqrt2$

Hence, option 2 is the correct option.

⇒ sec θ + tan θ + 1 = 0

⇒ sec θ + tan θ = -1

We know that,

⇒ sec² θ - tan² θ = 1

⇒ (sec θ + tan θ)(sec θ - tan θ) = 1

⇒ (-1)(sec θ - tan θ) = 1

⇒ sec θ - tan θ = -1

Hence, option 1 is the correct option.

⇒ sec θ + tan θ = x ....(1)

We know that,

⇒ sec² θ − tan² θ = 1

⇒ (sec θ + tan θ)(sec θ − tan θ) = 1

⇒ x (sec θ − tan θ) = 1

⇒ sec θ − tan θ = $\frac{1}{x}$ ....(2)

Adding eqn (1) and (2):

⇒ sec θ + tan θ + sec θ − tan θ = x + $\frac{1}{x}$

⇒ 2 sec θ = $x + \frac{1}{x}$

⇒ sec θ = $\frac{1}{2}\Big(x + \frac{1}{x}\Big)$

⇒ sec θ = $\frac{1}{2}\Big(\frac{x^2 + 1}{x}\Big)$

⇒ sec θ = $\frac{x^2 + 1}{2x}$

Hence, option 3 is the correct option.

⇒ sin θ − cos θ = 0

⇒ sin θ = cos θ

Divide by cos θ

⇒ $\frac{sin \theta}{\cos \theta}$ = 1

tan 45° = 1

Given expression,

(sin⁴ 45° + cos⁴ 45°)

$$\begin{aligned}\Rightarrow \Big(\frac{1}{\sqrt2}\Big)^4 + \Big(\frac{1}{\sqrt2}\Big)^4 \\[1em] \Rightarrow \Big(\frac{1}{4}\Big) + \Big(\frac{1}{4}\Big) \\[1em] \Rightarrow \frac{1}{2}.\end{aligned}$$

Hence, option 2 is the correct option.

⇒ a cot θ + b cosec θ = p

p² = (a cot θ + b cosec θ)²

p² = (a² cot² θ + b² cosec² θ + 2ab cot θ cosec θ)

⇒ b cot θ + a cosec θ = q

q² = (b cot θ + a cosec θ)²

q² = (b² cot² θ + a² cosec² θ + 2ab cot θ cosec θ)

⇒ p² − q² = (a² cot² θ + b² cosec² θ + 2ab cot θ cosec θ) - (b² cot² θ + a² cosec² θ + 2ab cot θ cosec θ)

= (a² cot² θ + b² cosec² θ + 2ab cot θ cosec θ - b² cot² θ - a² cosec² θ - 2ab cot θ cosec θ)

= (a² cot² θ + b² cosec² θ - b² cot² θ - a² cosec² θ )

= a² (cot² θ - cosec²) + b² (cosec² θ - cot² θ )

= a² (-1) + b² (1)

= b² - a².

Hence, option 2 is the correct option.

sec² θ + cosec² θ

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

Hence, option 1 is the correct option.

Given,

x = a cos³ θ and y = b sin³ θ

$$\begin{aligned}\Rightarrow \frac{x}{a} = \cos^3 θ \\[1em] \Rightarrow \Big(\frac{x}{a}\Big)^{\frac{2}{3}} = \cos^2 θ \\[1em] \Rightarrow \frac{y}{b} = \sin^3 θ \\[1em] \Rightarrow \Big(\frac{y}{b}\Big)^{\frac{2}{3}} = \sin^2 θ\end{aligned}$$

Add the expressions,

$\Big(\frac{x}{a}\Big)^{\frac{2}{3}} + \Big(\frac{y}{b}\Big)^{\frac{2}{3}}$ = cos² θ + sin² θ

$\Big(\frac{x}{a}\Big)^{\frac{2}{3}} + \Big(\frac{y}{b}\Big)^{\frac{2}{3}}$ = 1

Hence, option 3 is the correct option.

sin θ + cosec θ = 2

Let,

⇒ sin θ = x

⇒ cosec θ = $\frac{1}{x}$

$$\begin{aligned}\Rightarrow x + \frac{1}{x} = 2 \\[1em] \Rightarrow \frac{x^2 + 1}{x} = 2 \\[1em] \Rightarrow x^2 + 1 = 2x \\[1em] \Rightarrow x^2 + 1 - 2x = 0 \\[1em] \Rightarrow (x - 1)^2 = 0 \\[1em] \Rightarrow x = 1\end{aligned}$$

Hence,

sin θ = 1 and cosec θ = 1

⇒ sin³ θ + cosec³ θ

⇒ 1³ + 1³

⇒ 2

Hence, option 1 is the correct option.

Solving for option 3,

cos x + tan x sin x

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

Hence, option 3 is the correct option.

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

Hence, option 2 is the correct option.

Given,

sin A + sin² A = 1

It can be written as

sin A = 1 - sin² A …. (1)

We have to find the value of (cos² A + cos⁴ A)

Using the trigonometric identities,

cos² A = 1 - sin² A ….. (2)

From both the equations

sin A = cos² A

Now, (cos² A + cos⁴ A) = (cos² A + (sin A)²)

= cos² A + sin²A

cos² A + sin² A = 1

Therefore, (cos² A + cos4 A) = 1

Hence, option 2 is the correct option.

cos A + cos² A = 1

⇒ 1 - cos² A = cos A

⇒ sin² A = cos A

Given,

⇒ sin² A + sin⁴ A

⇒ sin² A + (sin² A)²

⇒ sin² A + cos² A [∵ sin² A = cos A]

⇒ 1.

Hence, option 1 is the correct option.

Rationalize the expression,

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

Hence, option 2 is the correct option.

Trigonometry identity :

sin² θ + cos² θ = 1

cosec² θ - cot² θ = 1

∴ Only statement (i) is correct.

Hence, Option 1 is the correct option.

Given,

⇒ sin² A + sin A = 1

⇒ sin A = 1 - sin² A

⇒ sin A = cos² A

Squaring both the sides, we get :

⇒ (sin A)² = (cos² A)²

⇒ sin² A = cos⁴ A ......................(1)

Solving L.H.S. of cos⁴ A + cos² A = 1, we get :

⇒ cos⁴ A + cos² A

⇒ sin² A + cos² A [From equation (1)]

⇒ 1.

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

∴ Assertion (A) is true.

As we know that, sin² A + cos² A = 1

⇒ 1 - sin² A = cos² A

∴ Reason (R) is true.

∴ Both (A) and (R) are true, and (R) is the correct explanation for (A).

Hence, option 1 is the correct option.

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

∴ Assertion (A) is true.

tan² θ + sec² θ = 1 is incorrect; the correct identity is sec² θ − tan² θ = 1.

∴ Reason (R) is false.

Hence, option 3 is the correct option.

Solving L.H.S,

(1 − cosec² θ)(1 − sec² θ)

= (− cot² θ)(− tan² θ) [∵ 1 − cosec² θ = − cot² θ and 1 − sec² θ = − tan² θ]

= $\frac{1}{\tan^2 \theta}$ (tan² θ) = 1

∴ Assertion (A) is true.

1 + tan² θ = sec² θ and 1 + cot² θ = cosec² θ are standard identities, and these are exactly the relations used to simplify the Assertion.

∴ Reason (R) is true and is the correct explanation of A.

Hence, option 1 is the correct option.

Given, sec θ + tan θ = p.

We know that sec² θ − tan² θ = 1.

⇒ (sec θ + tan θ)(sec θ − tan θ) = 1

⇒ (p)(sec θ − tan θ) = 1

⇒ sec θ − tan θ = $\frac{1}{p}$

Adding sec θ + tan θ = p and sec θ − tan θ = $\frac{1}{p}$ :

⇒ 2 sec θ = p + $\frac{1}{p}$ = $\frac{p^2 + 1}{p}$

⇒ sec θ = $\frac{p^2 + 1}{2p}$.

So the correct value is $\frac{p^2 + 1}{2p}$, not $\frac{2p}{p^2 + 1}$.

∴ Assertion (A) is false.

sec² θ − tan² θ = 1 is a standard Pythagorean identity.

∴ Reason (R) is true.

Hence, option 4 is the correct option.

Given, sin θ = cos θ.

⇒ $\frac{\sin \theta}{\cos \theta}$ = 1 ⇒ tan θ = 1, which for an acute angle gives θ = 45°.

Also, sin θ = cos θ ⇒ sin² θ = cos² θ, and since sin² θ + cos² θ = 1, we get 2 sin² θ = 1.

⇒ 2 sin² θ + tan² θ = 1 + (1)² = 1 + 1 = 2.

∴ Assertion (A) is true.

sin (90° − θ) = cos θ is a true complementary-angle identity.

∴ Reason (R) is true.

However, the value 2 sin² θ + tan² θ = 2 is obtained using tan θ = 1 and 2 sin² θ = 1, not from the complementary-angle relation. So R does not explain A.

Hence, option 2 is the correct option.

Given,

⇒ sec θ + tan θ = a

⇒ sec θ − tan θ = b

⇒ ab = (sec θ + tan θ)(sec θ - tan θ)

⇒ ab = sec² θ - tan² θ

⇒ ab = 1

So assertion (A) is true.

We know that,

⇒ sec² θ - tan² θ = 1

This is a fundamental trigonometric identity.

So reason (R) is true.

Thus, Both (A) and (R) are true and (R) is the correct explanation of (A).

Hence, option 1 is the correct option.