Question

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:$\sqrt{\frac{\text{1 + sin A}}{\text{1 - sin A }}}= \text{sec A+ tan A}$

Answer 1

$\sqrt{\frac{\text{1 + sin A}}{\text{1 - sin A }}}= \text{sec A+ tan A}$

LHS $= \sqrt{\frac{\text{1 + sin A}}{\text{(1 - sin A)}}}$

$⇒ \sqrt{\frac{\text{(1 + sin A)(1 + sin A)}}{\text{(1 - sin A)(1 + sin A)}}}$

$=\sqrt{\frac{ \text{(1 + sin A)²}}{\text{(1 - sin² A) }}}$ [a² - b² = (a - b)(a + b)]

$= \frac{\text{(1 + sinA)}}{\sqrt{\text{1 - sin² A}}}$

$=\frac{ \text{1 + sin A}}{\sqrt{\text{cos² A}}}$

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

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

$= \text{sec A + tan A}$

= R.H.S