Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\sqrt{\frac{\text{1 + sin A}}{\text{1 - sin A }}}= \text{sec A+ tan A}\)
Solution and Explanation
\(\sqrt{\frac{\text{1 + sin A}}{\text{1 - sin A }}}= \text{sec A+ tan A}\)
LHS \(= \sqrt{\frac{\text{1 + sin A}}{\text{(1 - sin A)}}}\)
\(⇒ \sqrt{\frac{\text{(1 + sin A)(1 + sin A)}}{\text{(1 - sin A)(1 + sin A)}}}\)
\(=\sqrt{\frac{ \text{(1 + sin A)²}}{\text{(1 - sin² A) }}}\) [identity: 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
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