Question

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

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

Answer 1

Answer: (A)

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