Trigonometrical Identities — Exercise 22(A)

Prove the following identity:

$\Big(\dfrac{\cosec A + 1}{\cosec A - 1}\Big) = \Big(\dfrac{1 + \sin A}{1 - \sin A}\Big)$

Answer

Solving L.H.S:

$\Rightarrow \dfrac{\cosec A + 1}{\cosec A - 1} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\sin A} + 1}{\dfrac{1}{\sin A} - 1} \\[1em] \Rightarrow \dfrac{\dfrac{1 + \sin A}{\sin A}}{\dfrac{1 - \sin A}{\sin A}} \\[1em] \Rightarrow \dfrac{(1 + \sin A) \times \sin A}{(1 - \sin A) \times \sin A} \\[1em] \Rightarrow \dfrac{1 + \sin A}{1 - \sin A}.$

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

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

Prove the following identity:

$\Big(\dfrac{\sec A - 1}{\sec A + 1}\Big) = \Big(\dfrac{1 - \cos A}{1 + \cos A}\Big)$

Answer

Solving L.H.S:

$\Rightarrow \dfrac{\sec A - 1}{\sec A + 1} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\cos A} - 1}{\dfrac{1}{\cos A} + 1} \\[1em] \Rightarrow \dfrac{\dfrac{1 - \cos A}{\cos A}}{\dfrac{1 + \cos A}{\cos A}} \\[1em] \Rightarrow \dfrac{(1 - \cos A) \times \cos A}{(1 + \cos A) \times \cos A} \\[1em] \Rightarrow \dfrac{1 - \cos A}{1 + \cos A}.$

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

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

Prove the following identity:

$\Big(\dfrac{\sin A \times \tan A}{1 - \cos A}\Big) = 1 + \sec A$

Answer

Solving L.H.S of the equation:

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

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

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

Prove the following identity:

$\Big(\dfrac{1}{\tan A + \cot A}\Big) = \cos A \times \sin A$

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{1}{\tan A + \cot A} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\sin A}{\cos A} + \dfrac{\cos A}{\sin A}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\sin^2 A + \cos^2 A}{\sin A \cos A}} \\[1em] \Rightarrow \dfrac{\sin A \cos A}{\sin^2 A + \cos^2 A} \\[1em] \Rightarrow \sin A \cos A$

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

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

Prove the following identity:

(1 + tan A)² + (1 - tan A)² = 2 sec² A

Answer

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.

Prove the following identity:

(sin² θ - 1) (tan² θ + 1) + 1 = 0

Answer

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.

Prove the following identity:

cosec A (1 + cos A)(cosec A - cot A) = 1

Answer

Solving L.H.S. of the equation :

⇒ cosec A(1 + cos A)(cosec A - cot A)

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

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

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

Prove the following identity:

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

Answer

Solving L.H.S. of the equation :

⇒ sec A(1 - sin A)(sec A + tan A)

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

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

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

Prove the following identity:

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

Answer

Solving L.H.S. of the equation :

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

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

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

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

Prove the following identity:

(cosec A + sin A)(cosec A - sin A) = cot² A + cos² A

Answer

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.

Prove the following identity:

(sec A + cos A)(sec A - cos A) = sin² A + tan² A

Answer

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.

Prove the following identity:

tan² A - sin² A = sin² A tan² A

Answer

Solving L.H.S of equation,

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

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

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

Prove the following identity:

cot² A - cos² A = cos² A cot² A

Answer

Solving L.H.S of equation,

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

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

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

Prove the following identity:

sec² A + cosec² A = sec² A cosec² A

Answer

Solving L.H.S of equation,

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

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

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

Prove the following identity:

tan² A + cot² A + 2 = sec² A cosec² A

Answer

Solving L.H.S of equation,

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

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

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

Prove the following identity:

sin A (1 + tan A) + cos A (1 + cot A) = sec A + cosec A

Answer

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

$\Rightarrow \sin A + \dfrac{\sin^2 A}{\cos A} + \cos A + \dfrac{\cos^2 A}{\sin A} \\[1em] \Rightarrow \sin A + \dfrac{\cos^2 A}{\sin A} + \cos A +\dfrac{\sin^2 A}{\cos A} \\[1em] \Rightarrow \dfrac{\sin^2 A + \cos^2 A}{\sin A} + \dfrac{\sin^2 A + \cos^2 A}{\cos A} \\[1em] \Rightarrow \dfrac{1}{\sin A} + \dfrac{1}{\cos A} \\[1em] \Rightarrow \sec A + \cosec A.$

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

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

Prove the following identity:

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

Answer

Solving L.H.S of equation,

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

By formula:

1 + tan² A = sec² A

1 + cot² A = cosec² A

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

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

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

Prove the following identity:

$\Big(\dfrac{1}{1 + \sin A}\Big) + \Big(\dfrac{1}{1 - \sin A}\Big) = 2 \sec^2 A$

Answer

Solving L.H.S. of the equation :

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

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

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

Prove the following identity:

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

Answer

Solving L.H.S,

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

Since,

L.H.S = R.H.S

Hence, proved that

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

Prove the following identity:

$\Big(\dfrac{\cosec A}{\cosec A - 1}\Big) + \Big(\dfrac{\cosec A}{\cosec A + 1}\Big) = 2 \sec^2 A$

Answer

Solving L.H.S. of the equation :

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

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

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

Prove the following identity:

(1 + cot A - cosec A)(1 + tan A + sec A) = 2

Answer

Solving L.H.S. of the equation :

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

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

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

Prove the following identity:

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

Answer

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

⇒ 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.

Prove the following identity:

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

Answer

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,

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

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

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

⇒ 1 - sin θ cos θ + sin θ cos θ

⇒ 1.

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

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

Prove the following identity:

$\Big(\dfrac{\tan A}{1 - \cot A}\Big) + \Big(\dfrac{\cot A}{1 - \tan A}\Big) = \sec A \cosec A + 1$

Answer

Solving L.H.S of equation,

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

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

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

Prove the following identity:

$\Big(\dfrac{\sin A}{1 + \cot A}\Big) - \Big(\dfrac{\cos A}{1 + \tan A}\Big) = \sin A - \cos A$

Answer

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

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

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

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

Prove the following identity:

$\Big(\dfrac{\tan \theta + \sin \theta}{\tan \theta - \sin \theta}\Big) = \Big(\dfrac{\sec \theta + 1}{\sec \theta - 1}\Big)$

Answer

Solving L.H.S of the equation,

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

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

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

Prove the following identity:

$\Big(\dfrac{\cot \theta + \cosec \theta - 1}{\cot \theta - \cosec \theta + 1}\Big) = \Big(\dfrac{1 + \cos \theta}{\sin \theta}\Big)$

Answer

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

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

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

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

Prove the following identity:

$\Big(\dfrac{\cot A - 1}{2 - \sec^2 A}\Big) = \Big(\dfrac{\cot A}{1 + \tan A}\Big)$

Answer

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

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

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

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

Prove the following identity:

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

Answer

The equation can be written as,

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

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

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

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

Hence, proved that

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

Prove the following identity:

$\Big(\dfrac{\sin A}{\cot A + \cosec A}\Big) = 2 + \Big(\dfrac{\sin A}{\cot A - \cosec A}\Big)$

Answer

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

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

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

$\Rightarrow \dfrac{\sin^2 A(\cos A − 1)}{\cos A + 1(\cos A − 1)} \\[1em] \Rightarrow \dfrac{\sin^2 A(\cos A − 1)}{\cos^2 A - 1} \\[1em] \Rightarrow \dfrac{\sin^2 A(\cos A − 1)}{-\sin^2 A} \\[1em] \Rightarrow -(\cos A − 1) \\[1em] \Rightarrow 1 - \cos A$

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

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

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

$\Rightarrow 2 + \dfrac{\sin^2 A(\cos A + 1)}{\cos A - 1(\cos A + 1)} \\[1em] \Rightarrow 2 + \dfrac{\sin^2 A(\cos A + 1)}{\cos^2 A - 1} \\[1em] \Rightarrow 2 + \dfrac{\sin^2 A(\cos A + 1)}{-\sin^2 A} \\[1em] \Rightarrow 2 -(\cos A + 1) \\[1em] \Rightarrow 1 - \cos A$

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

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

Prove the following identity:

$\cot A - \tan A = \Big(\dfrac{2 \cos^2 A - 1}{\sin A \cos A}\Big)$

Answer

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

$\Rightarrow \dfrac{\cos A}{\sin A} - \dfrac{\sin A}{\cos A} \\[1em] \Rightarrow \dfrac{\cos^2 A - \sin^2 A}{\sin A \cos A} \\[1em] \Rightarrow \dfrac{\cos^2 A - (1 - \cos^2 A)}{\sin A \cos A} \\[1em] \Rightarrow \dfrac{\cos^2 A - 1 + \cos^2 A}{\sin A \cos A} \\[1em] \Rightarrow \dfrac{2\cos^2 A - 1}{\sin A \cos A}.$

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

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

Prove the following identity:

$\sqrt{\Big(\dfrac{1 + \cos A}{1 - \cos A}\Big)} = \cosec A + \cot A$

Answer

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

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

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

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

Prove the following identity:

$\sqrt{\Big(\dfrac{1 - \sin A}{1 + \sin A}\Big)} = \sec A - \tan A$

Answer

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

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

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

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

Prove the following identity:

$\Big(\dfrac{\cot^2 A}{(\cosec A + 1)^2}\Big) = \Big(\dfrac{1 - \sin A}{1 + \sin A}\Big)$

Answer

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

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

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

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

Prove the following identity:

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

Answer

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

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