Question

If $y=sin^{2}\,(cot^{-1} \sqrt{\frac{1+x}{1-x}})$, then $\frac{dy}{dx}$ is equal to

• A. $2 \,sin \,2x$
• B. $sin\, 2x$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$
• E. $cos\,2x$

Answer 1

Answer: (D)

$y=\sin ^{2}\, \cot ^{-1} \sqrt{\frac{1+x}{1-x}}$
Put, $ x=\cos \,\theta$
$\Rightarrow \, \theta=\cos ^{-1} x\,\,\,\,\,\dots(i)$
$y=\sin ^{2} \cot ^{-1} \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$
$y=\sin ^{2} \cot ^{-1} \sqrt{\frac{2 \cos ^{2} \theta / 2}{2 \sin ^{2} \theta / 2}}$
$y=\sin ^{2} \cot ^{-1}(\cot \theta / 2)$
$y=\sin ^{2} \theta / 2$
$y=\frac{1-\cos \theta}{2}$
$y=\frac{1-x}{2}$
[From Eq. (i)]
Differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=-\frac{1}{2}$