Question

$\frac{d}{dx}\left[{\sin }^2\left({\cot }^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right)\right)\right]=$

• A. $0$
• B. $-\frac{1}{2}$
• C. $\frac{1}{2}$
• D. $-1$

Answer 1

The correct option is B

$-\frac{1}{2}$

Simplifying the function using Trigonometric identities

Given, $y=\left[{\sin }^2\left({\cot }^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right)\right)\right]$

Let,

$\begin{array}{rcl}\left(\sqrt{\frac{1+x}{1-x}}\right)& =& \cot \left(\theta \right)\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{1+x}{1-x}& =& {\cot }^2\left(\theta \right)\end{array}$

$\Rightarrow$ $\begin{array}{rcl}x& =& \frac{{\cot }^2\left(\theta \right)-1}{{\cot }^2\left(\theta \right)+1}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}x& =& \frac{{\cos }^2\left(\theta \right)-{\sin }^2\left(\theta \right)}{{\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}x& =& \cos \left(2\theta \right)\end{array}$

As,

$\begin{array}{rcl}\left(\sqrt{\frac{1+x}{1-x}}\right)& =& \cot \left(\theta \right)\end{array}$

$\Rightarrow$ $\begin{array}{rcl}y& =& {\sin }^2\left(\theta \right)\end{array}$

$\begin{array}{rcl}\therefore \frac{dy}{dx}& =& 2\sin \theta \cos \theta \times \left(\frac{-1}{2\sin \left(2\theta \right)}\right)\\ & =& \mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\end{array}$

Hence, Option (B) is the correct answer.