Question

If $2y={\left({\cot }^{-1}\left(\frac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x}\right)\right)}^2,x\in (0,\frac{\pi }{2})$ then $\frac{dy}{dx}$ is equal to:

• A. $\frac{\pi }{6}-x$
• B. $\frac{\pi }{3}-x$
• C. $x-\frac{\pi }{6}$
• D. $2x-\frac{\pi }{3}$

Answer 1

The correct option is C $x-\frac{\pi }{6}$

$2y={\left({\cot }^{-1}\left(\frac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x}\right)\right)}^2$
$\Rightarrow 2y={\left({\cot }^{-1}\left(\frac{\sqrt{3}+\tan x}{1-\sqrt{3}\tan x}\right)\right)}^2$
$\Rightarrow 2y={\left({\cot }^{-1}\left(\tan (\frac{\pi }{3}+x)\right)\right)}^2$
$\Rightarrow 2y={(\frac{\pi }{2}-{\tan }^{-1}\left(\tan (\frac{\pi }{3}+x)\right))}^2$
$\Rightarrow 2y={(\frac{\pi }{2}-(\frac{\pi }{3}+x))}^2$
$\Rightarrow 2y={(x-\frac{\pi }{6})}^2$
$\Rightarrow 2y=x^2-\frac{\pi }{3}x+\frac{{\pi }^2}{36}$
$\therefore y^{\prime }=x-\frac{\pi }{6}$