Question

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

• A. $2sin2x$
• B. $sin2x$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$
• E. $cos2x$

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
$\frac{dy}{dx}=-\frac{1}{2}$