Question

✓1+sinA÷1-sinA=secA+tanA​

Answer 1

Answer:

$LHS =\sqrt{\frac{1+sinA}{1-sinA} }$

$= \sqrt{\frac{1+sinA}{1-sinA} X \frac{1+sinA}{1+sinA} }$

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

$= (\sqrt{\frac{1+sinA}{cosA} })^2$

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

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

$= secA + tanA$

$= RHS$

Hence Proved.

Do mark as brainliest if it helped!