Question

$\frac {d}{dx} \left[ {Cos^2} \left(Cot^{-1} \sqrt{\frac{2+x}{2-x}} \right) \right]$ is ______

• A. $\frac {1}{4}$
• B. $\frac {1}{2}$
• C. $\frac {-1}{2}$
• D. $\frac {-3}{4}$

Answer 1

Answer: (A)

$\frac{d}{d x}\left\{\cos ^{2}\left(\cot ^{-1} \sqrt{\frac{2+x}{2-x}}\right)\right\}$ put $x=2 \cos \theta$...(i)
$\Rightarrow \frac{d}{d x}\left\{\cos ^{2}\left(\cot ^{-1} \sqrt{\frac{2+\cos \theta \cdot 2}{2-\cos \theta \cdot 2}}\right)\right\}$
$=\frac{d}{d x}\left\{\cos ^{2}\left(\cot ^{-1} \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\right)\right\}$
$=\frac{d}{d x}\left\{\cos ^{2}\left(\cot ^{-1} \sqrt{\frac{2 \cos ^{2} \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}}}\right)\right\}$
$=\frac{d}{d x}\left\{\cos ^{2}\left(\cot ^{-1}(\cot \theta / 2)\right)\right\}$
$=\frac{d}{d x}\left\{\cos ^{2} \frac{\theta}{2}\right\}=\frac{1}{2} \cdot \frac{d}{d x}(1+\cos \theta)$
$=\frac{d}{d x}\left(\frac{1}{2}+\frac{1}{2} \cdot \frac{x}{2}\right)$
$=\frac{d}{d x}\left(\frac{1}{2}+\frac{x}{4}\right)$ [from Eq. (i)]
$=(0+1 / 4)=1 / 4$