Question

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

Answer 1

Consider the following equation.

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

$=\sqrt{\frac{{(1-\sin x)}^2}{{\cos }^{2}x}}$

$=\frac{1-\sin x}{\cos x}$

$=\sec x-\tan x$

Hence, this is the correct answer.