Question

$\frac{\pi }{2}<x<\frac{3\pi }{2}$, then $X=\frac{1}{\cos x-\sqrt{{\tan }^{2}x-{\sin }^{2}x}}=$

• A. $cosecx$
• B. $\sec x$
• C. $\sin x$
• D. $\cos x$

Answer 1

Answer: (D)

$\cos x$

Given expression is $\frac{1}{\cos x-\sqrt{{\tan }^{2}x-{\sin }^{2}x}}$

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

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

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

$=\frac{1}{\cos x+\sin x\tan x}$

$=\frac{1}{\cos x+\frac{{\sin }^{2}x}{\cos x}}$

$=\cos x$