Question

Hello friend solve this problem

Answer 1

It was the proof of this question

Answer 2

$Given : \sqrt{\frac{1 + sinA}{1 - sinA}}$

$= > \sqrt{\frac{1 + sinA}{1 - sinA}} * \sqrt{\frac{1 + sinA}{1 + sinA}}$

$= > \frac{\sqrt{(1 + sinA) * (1 + sinA)}}{\sqrt{(1 - sinA) * (1 + sinA)}}$

$= > \frac{\sqrt{(1 + sinA)^2}}{\sqrt{1 - sin^2A}}$

$= > \frac{\sqrt{1 + sinA)^2}}{\sqrt{cos^2A}}$

$= > \frac{1 + sinA}{cosA}$

$= > \frac{1}{cosA} + \frac{sinA}{cosA}$

= > secA + tanA

Hope it helps!