Question

If $x\in (\frac{-\pi }{2},\frac{\pi }{2}),$ then $\sqrt{\frac{1-\sin x}{1+\sin x}}$ is equal to

• A. $\sec x-\tan x$
• B. $\sec x+\tan x$
• C. $\tan x-\sec x$
• D. $-\tan x-\sec x$

Answer 1

The correct option is A $\sec x-\tan x$

$\sqrt{\frac{1-\sin x}{1+\sin x}}=\sqrt{\frac{(1-\sin x{)}^2}{1-{\sin }^{2}x}}=\sqrt{\frac{(1-\sin x{)}^2}{{\cos }^{2}x}}=\frac{|(1-\sin x)|}{|\cos x|}=\frac{1-\sin x}{|\cos x|}(\because 1-\sin x\ge 0)$

When $x\in (-\frac{\pi }{2},\frac{\pi }{2}),$ then $|\cos x|=\cos x$, so
$\sqrt{\frac{1-\sin x}{1+\sin x}}=\sec x-\tan x$