Question

Find the value of the expression $\frac{\sqrt{1-{\cos }^{2}x}}{\sin x}+\frac{\sqrt{1-{\sin }^{2}x}}{\cos x}$, given that $0<x<\frac{\pi }{2}$.

• A. $0$
• B. $1$
• C. $2$
• D. $-\tan x$
• E. $\sin 2x$

Answer 1

Answer: (C)

$2$

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

Therefore, Answer is $2$