If 2y = {left( {{{cot }^{ - 1}}left( {frac{{s | vedclass

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(A) Given $2y = {\left( {{{\cot }^{ - 1}}\left( {\frac{{\sqrt 3 \cos x + \sin x}}{{\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$.Divide numerator an

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $2y = {\left( {{{\cot }^{ - 1}}\left( {\frac{{\sqrt 3 \cos x + \sin x}}{{\cos x - \sqrt 3 \sin x}}} ight)} ight)^2}$.Divide numerator and denominator by $2$ inside the bracket:$\frac{{\frac{{\sqrt 3 }}{2}\cos x + \frac{1}{2}\sin x}}{{\frac{1}{2}\cos x - \frac{{\sqrt 3 }}{2}\sin x}} = \frac{{\sin(x + \frac{\pi }{3})}}{{\cos(x + \frac{\pi }{3})}} = 	an(x + \frac{\pi }{3})$.Thus,$2y = {\left( {{{\cot }^{ - 1}}(	an(x + \frac{\pi }{3}))} ight)^2}$.Using ${{\cot }^{ - 1}}(	an 	heta ) = \frac{\pi }{2} - 	heta$:$2y = {\left( {\frac{\pi }{2} - (x + \frac{\pi }{3})} ight)^2} = {\left( {\frac{\pi }{6} - x} ight)^2}$.Differentiating with respect to $x$:$2\frac{{dy}}{{dx}} = 2(\frac{\pi }{6} - x) \cdot (-1)$.$\frac{{dy}}{{dx}} = -(\frac{\pi }{6} - x) = x - \frac{\pi }{6}$.

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

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