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  4. If 2y =( cot-1 ((√3 cos x + sin x/ cos x -√3 sin x)))2 , x ∈(0, (π/2)) , then (dy/dx) is equal to :

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Q. 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 :

JEE MainJEE Main 2019Continuity and Differentiability

Solution:

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{cases} \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{cases}$
$ \therefore \; \; 2y = \begin{cases} \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{cases}$
$ \therefore 2 \frac{dy}{d} = \begin{cases} 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{cases}$