Question

solve it fast everyone!!!!!!​

Answer 1

$\large\underline{\sf{Given \:Question - }}$

$\sf \: 3\pi < x < 4\pi, \: then \: \dfrac{d}{dx}\bigg( \: {tan}^{ - 1} \sqrt{\dfrac{1 + cos\dfrac{x}{2} }{1 - cos\dfrac{x}{2} } \: \: } \bigg)$

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

Let assume that

$\rm :\longmapsto\:y = \: {tan}^{ - 1} \sqrt{\dfrac{1 + cos\dfrac{x}{2} }{1 - cos\dfrac{x}{2} } \: \: }$

We know that

$\boxed{ \tt{ \: 1 + cos2x = {2cos}^{2}x \: }} \\ \\ and \\ \\ \boxed{ \tt{ \: 1 - cos2x = {2sin}^{2}x \: }}$

So, using this identity, we get

$\rm :\longmapsto\:y = {tan}^{ - 1} \sqrt{\dfrac{ {2cos}^{2} \dfrac{x}{4}}{ {2sin}^{2}\dfrac{x}{4}}}$

$\rm :\longmapsto\:y = {tan}^{ - 1} \sqrt{\dfrac{ {cos}^{2} \dfrac{x}{4}}{ {sin}^{2}\dfrac{x}{4}}}$

$\rm :\longmapsto\:y = {tan}^{ - 1} \sqrt{ {cot}^{2} \dfrac{x}{4}}$

$\rm \implies\:y = {tan}^{ - 1}\bigg |cot\dfrac{x}{4}\bigg|$

As, it is given that,

$\red{\rm :\longmapsto\:3\pi < x < 4\pi \: \rm \implies\:\dfrac{3\pi}{4} < \dfrac{x}{4} < \pi \: }$

$\red{\rm \implies\:\dfrac{x}{4} \: lies \: in \: second \: quadrant}$

$\red{\rm \implies\:cot\dfrac{x}{4} < 0}$

So, it implies

$\rm :\longmapsto\:y = - {tan}^{ - 1}\bigg(cot\dfrac{x}{4} \bigg)$

$\rm :\longmapsto\:y = - {tan}^{ - 1}\bigg(tan\bigg[\dfrac{\pi}{2} - \dfrac{x}{4}\bigg]\bigg)$

$\rm :\longmapsto\:y = - \bigg[\dfrac{\pi}{2} - \dfrac{x}{4}\bigg]$

$\rm :\longmapsto\:y = - \dfrac{\pi}{2} + \dfrac{x}{4}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}\bigg[ - \dfrac{\pi}{2} + \dfrac{x}{4}\bigg]$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}k = 0 \: }}$

and

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

So, using this, we get

$\rm \implies\:\boxed{ \tt{ \: \dfrac{dy}{dx} = \frac{1}{4} \: }}$

Hence,

$\boxed{\sf \:3\pi <x <4\pi, \: then\:\dfrac{d}{dx}\bigg({tan}^{ - 1}\sqrt{\dfrac{1 + cos\dfrac{x}{2} }{1-cos\dfrac{x}{2}}} \bigg) = \frac{1}{4}}$

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Additional Information :-

$\red{\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x}\\ \\ \sf {sin}^{ - 1}x & \sf \dfrac{1}{ \sqrt{1 - {x}^{2} } } \\ \\ \sf {cos}^{ - 1}x & \sf \dfrac{ - 1}{ \sqrt{1 - {x}^{2} } } \\ \\ \sf {tan}^{ - 1}x & \sf \dfrac{1}{1 + {x}^{2} } \\ \\ \sf {sec}^{ - 1}x & \sf \dfrac{1}{x \sqrt{ {x}^{2} - 1 } } \end{array}} \\ \end{gathered}}$