# If $2y=\left ( \cot ^{-1}\left ( \frac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x} \right ) \right )^{2},x\; \epsilon \left ( 0,\frac{\pi}{2} \right )$ then $\frac{dy}{dx}$ is equal to : Option 1) $\frac{\pi }{6}-x$ Option 2) $\frac{\pi }{3}-x$ Option 3) $x-\frac{\pi }{6}$   Option 4) $2x-\frac{\pi }{3}$

$2y=\left ( \cot ^{-1}\left ( \frac{\sqrt{3}\cos x+\sin x}{\cos x-\sqrt{3}\sin x} \right ) \right )^{2} \; \; \; \; \; \; x\; \epsilon \left ( 0, \frac{\pi }{2}\right )$

then $\frac{dy}{dx}$,

$2y=\left ( \cot ^{-1}\left ( \frac{\frac{\sqrt{3}\cos x}{\cos x}+\frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x}-\frac{\sqrt{3}\sin x}{\cos x}} \right ) \right )^{2}$

$=\left ( \cot ^{-1}\left ( \frac{\sqrt{3}+\tan x}{1-\sqrt{3}\tan x} \right ) \right )^{2}$

$=\left ( \cot ^{-1}\left ( \tan \left ( \frac{\pi}{3}+x \right ) \right ) \right )^{2}$

$=\left ( \frac{\pi}{2}-\tan ^{-1}\left ( \tan \left ( \frac{\pi}{3}+x \right ) \right ) \right )^{2}$

$=\left ( \frac{\pi}{2}-\left ( \frac{\pi}{3}+x \right ) \right )^{2}$

$= \left ( \frac{\pi}{2}-\frac{\pi}{3}-x \right )^{2}=\left ( \frac{\pi}{6}-x \right )^{2}$

$2y=\left ( x-\frac{\pi}{6} \right )^{2}$

$\frac{dy}{dx}=\frac{1}{2}\times2\left ( x-\frac{\pi}{6} \right )$

$=x-\frac{\pi}{6}$

Option 1)

$\frac{\pi }{6}-x$

Option 2)

$\frac{\pi }{3}-x$

Option 3)

$x-\frac{\pi }{6}$

Option 4)

$2x-\frac{\pi }{3}$

Exams
Articles
Questions