Question

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

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

Answer 1

Answer: (D)

Consider ${\cot }^{-1}\left(\frac{\frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x}{\frac{1}{2}\sin x-\frac{\sqrt{3}}{2}\sin x}\right)$
$={\cot }^{-1}\left(\frac{\sin \left(x+\frac{\pi }{3}\right)}{\cos \left(x+\frac{\pi }{3}\right)}\right)$
$={\cot }^{-1}\left(\tan \left(x+\frac{\pi }{3}\right)\right)=\frac{\pi }{2}-{\tan }^{-1}\left(\tan \left(x+\frac{\pi }{3}\right)\right)$
$=\{\begin{array}{l}\frac{\pi }{2}-\left(x+\frac{\pi }{3}\right)=\left(\frac{\pi }{6}-x\right);0<x<\frac{\pi }{6}\\ \frac{\pi }{2}-\left(\left(x-\frac{\pi }{3}\right)-\pi \right)=\left(\frac{7\pi }{6}-x\right);\frac{\pi }{6}<x<\frac{\pi }{2}\end{array}$
$\therefore 2y=\{\begin{array}{l}{\left(\frac{\pi }{6}-x\right)}^2;0<x<\frac{\pi }{6}\\ {\left(\frac{7x}{6}-x\right)}^2;\frac{\pi }{6}<x<\frac{\pi }{2}\end{array}$
$\therefore 2\frac{dy}{d}=\{\begin{array}{l}2\left(\frac{\pi }{6}-x\right).\left(-1\right);0<x<\frac{\pi }{6}\\ 2\left(\frac{7\pi }{6}-x\right).\left(-1\right);\frac{\pi }{6}<x<\frac{\pi }{2}\end{array}$