Question

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:$\frac{(\text{cos A - sin A + 1})}{\text{(cos A + sin A + 1)}} = \text{cosec A + cot A}$, using the identity cosec2 A = 1 + cot2 A.

Answer 1

$\frac{(\text{cos A - sin A + 1})}{\text{(cos A + sin A + 1)}} = \text{cosec A + cot A}$

L.H.S = $\frac{(\text{cos A - sin A + 1})}{\text{(cos A + sin A + 1)}}$
$=\frac{\left [\frac{\text{cos A}}{\text{sin A}} -\frac{ \text{sin A}}{\text{sin A }}+ \frac{1}{\text{sin A}}\right]}{\left[\frac{\text{cos A}}{\text{sin A}} +\frac{\text{ sin A}}{\text{sin A}} - \frac{1}{\text{sin A}}\right]}$

$⇒ \frac{\text{(cot A - 1 + cosec A) }}{\text{ (cot A + 1 - cosec A)}}$

$⇒ \frac{\text{cot A - (1 - cosec A) }}{\text{ cot A+ (1 - cosec A)}}$

multiplying $\text{[cot A - (1 - cosec A)]}$ in numerator and denominator

$= \frac{\text{[(cot A) - (1 - cosec A) × (cot A) - (1 - cosec A)]}}{\text{[(cot A) + (1 - cosec A)× (cot A) - (1 - cosec A)]}}$

$=\frac{\text{ [cot A - (1 - cosec A)]²}}{\text{[(cot A)² - (1 - cosecA)²]}}$

$= \frac{\text{[cot² A + (1 -cosecA)² - 2cot A(1 - cosecA)]}}{\text{[cot² A - (1 + cosec² A - 2cosec A)]}}$

$=\frac{\text{ (cot² A + 1 + cosec² A - 2cosec A - 2cot A + 2cot A cosec A)}}{\text{(cot²A - (1 + cosec² A - 2cosec A))}}$

$= \frac{\text{(2cosec² A+ 2cot A cosec A - 2cot A - 2cosecA)}}{\text{(cot² A - 1 - cosec² A + 2cosec A)}}$

$=\frac{\text{ 2cosec A(cosec A+ cot A) - 2(cot A + cosec A)}}{\text{(cot² A - cosec² A - 1 + 2cosec A)}}$

$=\frac{\text{ (cosec A + cot A)(2cosec A - 2)}}{(\text{- 1 - 1 + 2cosec A)}}$

$=\frac{\text{ (cosec A + cot A)(2cosec A - 2)}}{\text{(2cosec A - 2)}}$

$=\text{ cosec A + cot A}$

= R.H.S