Question

$\sf if\: 2y = \bigg(cot^{-1}\bigg(\dfrac{\sqrt{3}cosx+sinx}{cosx-\sqrt{3}sinx}\bigg)\bigg)^2,x\in\bigg(0,\dfrac{\pi}{2}\bigg)$
Then dy/dx is equal to

Answer 1

$\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \bf{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{ n- 1} }}$

$\boxed{ \bf{ \: \dfrac{d}{dx}x = 1}}$

$\boxed{ \bf{ \: {tan}^{ - 1} x + {cot}^{ - 1}x = \dfrac{\pi}{2} }}$

$\boxed{ \bf{ \: {tan}^{ - 1}x + {tan}^{ - 1} y = {tan}^{ - 1} \bigg( \dfrac{x + y}{1 - xy} \bigg) }}$

$\boxed{ \bf{ \: {tan}^{ - 1} (tanx) = x}}$

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\sf \: 2y = \bigg(cot^{-1}\bigg(\dfrac{\sqrt{3}cosx+sinx}{cosx-\sqrt{3}sinx}\bigg)\bigg)^2$

We know,

$\boxed{ \bf{ \: {cot}^{ - 1} x = \dfrac{\pi}{2} - {tan}^{ - 1} x}}$

So,

Given expression can be rewritten as

$\rm :\longmapsto\: 2y = \bigg(\dfrac{\pi}{2} - tan^{-1}\bigg(\dfrac{\sqrt{3}cosx+sinx}{cosx-\sqrt{3}sinx}\bigg)\bigg)^2$

$\rm :\longmapsto\: 2y = \bigg(\dfrac{\pi}{2} - tan^{-1}\bigg(\dfrac{\sqrt{3} \: \dfrac{cosx}{cosx} +\dfrac{sinx}{cosx} }{1-\sqrt{3} \times \dfrac{sinx}{cosx} }\bigg)\bigg)^2$

$\rm :\longmapsto\: 2y = \bigg(\dfrac{\pi}{2} - tan^{-1}\bigg(\dfrac{\sqrt{3}+tanx}{1-\sqrt{3}tanx}\bigg)\bigg)^2$

Now,

We know that

$\boxed{ \bf{ \: {tan}^{ - 1} \bigg(\dfrac{x + y}{1 - xy} \bigg) = {tan}^{ - 1}x + {tan}^{ - 1}y}}$

Therefore,

$\rm :\longmapsto\: 2y = \bigg(\dfrac{\pi}{2} - \bigg( tan^{-1} \sqrt{3} + {tan}^{ - 1} (tanx) \bigg)\bigg)^2$

$\rm :\longmapsto\: 2y = \bigg(\dfrac{\pi}{2} - \dfrac{\pi}{3} - x\bigg)^2$

$\rm :\longmapsto\: 2y = \bigg(\dfrac{\pi}{6} -x \bigg)^2$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\: \dfrac{d}{dx} (2y) =\dfrac{d}{dx} \bigg(\dfrac{\pi}{6} -x \bigg)^2$

$\rm :\longmapsto\: \cancel2\dfrac{dy}{dx} = \cancel2 \bigg(\dfrac{\pi}{6} -x \bigg)\dfrac{d}{dx}\bigg(\dfrac{\pi}{6} -x \bigg)$

$\rm :\longmapsto\: \dfrac{dy}{dx} = \bigg(\dfrac{\pi}{6} -x \bigg)(0 - 1)$

$\bf\implies \:\dfrac{dy}{dx} = x - \dfrac{\pi}{6}$

Answer 2

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