# if 2y =

Question:

If $2 y=\left(\cot ^{-1}\left(\frac{\sqrt{3} \cos x+\sin x}{\cos x-\sqrt{3} \sin x}\right)\right)^{2}, x \in\left(0, \frac{\pi}{6}\right)$ then $\frac{d y}{d x}$

is equal to :

1. (1) $\frac{\pi}{6}-x$

2. (2) $x-\frac{\pi}{6}$

3. (3) $\frac{\pi}{3}-x$

4. (4) $2 x-\frac{\pi}{3}$

Correct Option: , 2

Solution:

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

$\Rightarrow 2 y=\left[\cot \left(\frac{\cos \left(\frac{\pi}{\sin (-)}\right)}{\sin (-)}\right)\right]$

$\Rightarrow 2 y=\left[\cot ^{-1}\left(\cot \left(\frac{\pi}{6}-x\right)\right)\right]^{2} \because \frac{\pi}{6}-\in\left(-\frac{\pi}{3} \frac{\pi}{6}\right)$

$\Rightarrow 2 y= \begin{cases}\left(\frac{7 \pi}{6}-x\right)^{2}, & \text { if } \frac{\pi}{6}-x \in\left(\frac{-\pi}{3}, 0\right) \\ \left(\frac{\pi}{6}-x\right)^{2}, & \text { if } \frac{\pi}{6}-x \in\left(0, \frac{\pi}{6}\right)\end{cases}$

$\Rightarrow \frac{d y}{d x}= \begin{cases}x-\frac{7 \pi}{6} & \text { if } x \in\left(\frac{\pi}{6}, \frac{\pi}{2}\right) \\ x-\frac{\pi}{6} & \text { if } x \in\left(0, \frac{\pi}{6}\right)\end{cases}$