Question

Prove that :
$\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A;A<90°$.

Answer 1

To prove:- $\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A$

Proof:-

$\sqrt{\frac{1+\sin A}{1-\sin A}}$

$=\frac{\left(\frac{1}{\sqrt{\cos A}}\right)}{\left(\frac{1}{\sqrt{\cos A}}\right)}\times \sqrt{\frac{1+\sin A}{1-\sin A}}$

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

$=\sqrt{\frac{\sec A+\tan A}{\sec A-\tan A}}\times \sqrt{\frac{\sec A+\tan A}{\sec A+\tan A}}$

$=\sqrt{\frac{{(\sec A+\tan A)}^2}{{\sec }^{2}A-{\tan }^{2}A}}$

$=\frac{\sqrt{{(\sec A+\tan A)}^2}}{1}(\because 1+{\tan }^{2}A={\sec }^{2}A)$

$=\sec A+\tan A$

Hence proved.