Question

If $\frac{\pi }{2}<x<\frac{3\pi }{2}$, 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. $noneofthese$

Answer 1

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

Rationalizing the denominator of $\sqrt{\frac{1-\sin x}{1+\sin x}}$.

$\sqrt{\frac{1-\sin x}{1+\sin x}}=\sqrt{\frac{\left(1-\sin x\right)\left(1-\sin x\right)}{\left(1+\sin x\right)\left(1-\sin x\right)}}$

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

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

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

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

$=\sec x-\tan x$