Question

If $y={\tan }^{-1}\sqrt{\frac{1-\cos x}{1+\cos x}}$, then for $0<x<\frac{\pi }{2}$, $\frac{dy}{dx}=$

• A. $2{\sec }^2\left(x/2\right)$
• B. $\frac{1}{2}{\sec }^2\left(x/2\right)$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}{\sec }^2\left(x/2\right)$

Answer 1

The correct option is C $\frac{1}{2}$

We know that $\cos 2x=1-2{\sin }^{2}x=2{\cos }^{2}x-1$

Using above identity, we have
$1-\cos x=2{\sin }^2\frac{x}{2}$

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

Using these values in given equation we get
$\tan y=\tan \frac{x}{2}\Rightarrow y=\frac{x}{2}\Rightarrow y^{\prime }=\frac{1}{2}$