Question

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

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

Answer 1

Proof:

Consider the LHS side of the given expression.

$$\begin{gathered} \mathrm{LHS}=\sqrt{\frac{1+\sin A}{1-\sin A}} \\ =\sqrt{\frac{{\displaystyle \frac{1+\sin A}{\cos A}}}{{\displaystyle \frac{1-\sin A}{\cos A}}}}\left[\mathrm{Divide}\mathrm{numerator}\mathrm{and}\mathrm{dnominator}\mathrm{by}\cos A\right] \\ =\sqrt{\frac{secA+\tan A}{secA-\tan A}}\left[\because \frac{1}{\cos A}=\sec A\mathrm{and}\frac{\sin A}{\cos A}=\tan A\right] \\ =\sqrt{\frac{secA+\tan A}{secA-\tan A}}\times \sqrt{\frac{secA+\tan A}{secA+\tan A}}\left[\mathrm{Rationalization}\right] \\ =\sqrt{\frac{{(secA+\tan A)}^2}{se{c}^{2}A-{\tan }^{2}A}} \\ =\sqrt{\frac{{(secA+\tan A)}^2}{1+{\tan }^{2}A-{\tan }^{2}A}}\left[\mathrm{Use}1+{\tan }^{2}A={\sec }^{2}A\right] \\ =\sec A+\tan A \\ =\mathrm{RHS} \end{gathered}$$

Hence proved.