Question

If $f\left(x\right)={\sin }^{-1}\left(\frac{1-\cos 2x}{2\sin x}\right)$, then $\left|f^{{}^{\prime }}\left(x\right)\right|$ is equal to

• A. $\left|\sin x\right|$
• B. $x$
• C. $0$
• D. $\left|\cos x\right|$
• E. $1$

Answer 1

The correct option is A $\left|\sin x\right|$

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

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

Therefore, $f\left(x\right)=\left|x\right|=\pm x$

$\Rightarrow f^{{}^{\prime }}\left(x\right)=\pm 1$

$\Rightarrow \left|f^{{}^{\prime }}\left(x\right)\right|=1$