Question

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

• A. $\frac{-1}{2}$
• B. $\frac{1}{ 1 + x }$
• C. $\frac{1}{1 -x}$
• D. $1$

Answer 1

Answer: (A)

$\frac{-1}{2}$

Answer 2

Given, $y=\sin ^{2}\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$
Let $x =\cos 2 \theta \dots$(i)
$\therefore y =\sin ^{2}\left[\cot ^{-1} \sqrt{\frac{1+\cos 2 \theta}{1-\cos 2 \theta}}\right]$
$\left[\because 1+\cos 2 \theta=2 \cos ^{2} \theta \text { and } 1-\cos 2 \theta=2 \sin ^{2} \theta\right]$
$= \sin ^{2}\left[\cot ^{-1} \sqrt{\frac{2 \cos ^{2} \theta}{2 \sin ^{2} \theta}}\right]$
$= \sin ^{2}\left[\cot ^{-1}(\cot \theta)\right]$
$\Rightarrow y=\sin ^{2} \theta$
[Differentiating w.r.t. $\theta$, we get $]$
$\Rightarrow \frac{d y}{d \theta}=2 \sin \theta \cos \theta$
$\Rightarrow \frac{d y}{d \theta}=\sin 2 \theta \ldots$(ii)
Differentiating E (i) w.r.t. $\theta$, we get
$\frac{d x}{d \theta}=-2 \sin 2 \theta \dots$(iii)
$\because \frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{\sin 2 \theta}{-2 \sin 2 \theta}$
[by Eqs. (ii) and (iii)]
$\Rightarrow \frac{d y}{d x}=-\frac{1}{2}$