If A and B are complementary angles, prove that :

(i) cot B + cos B = sec A cos B (1 + sin B)

(ii) cot A cot B - sin A cos B - cos A sin B = 0

(iii) cosec² A + cosec² B = cosec² A cosec² B

(iv) $\dfrac{\text{sin A + sin B}}{\text{sin A - sin B}} + \dfrac{\text{cos B - cos A}}{\text{cos B + cos A}} = \dfrac{2}{\text{2 sin}^2 A - 1}$

Given,

A + B = 90°

B = 90° - A and A = 90° - B.

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

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

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 $\times \dfrac{1}{\text{cot A}}$ - (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

⇒ $\dfrac{1}{\text{sin}^2 A}$ + sec² A

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

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

By formula,

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

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

⇒ 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 :

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

By formula,

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

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

By formula,

cos² A = 1 - sin² A

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

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

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