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)

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 $]$
$\Rightarrow \frac{dy}{d\theta }=2\sin \theta \cos \theta$
$\Rightarrow \frac{dy}{d\theta }=\sin 2\theta \dots$(ii)
Differentiating Eq. (i) w.r.t. $\theta$, we get
$\frac{dx}{d\theta }=-2\sin 2\theta \dots$(iii)
$\because \frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{\sin 2\theta }{-2\sin 2\theta }$
[by Eqs. (ii) and (iii)]
$\Rightarrow \frac{dy}{dx}=-\frac{1}{2}$