Question

$\sf if \: y(\alpha)=\sqrt{2\bigg(\dfrac{tan\alpha+cot\alpha}{1+tan^2\alpha}\bigg) +\dfrac{1}{sin^2\alpha} } , \alpha \in \bigg(\dfrac{3\pi}{4},\pi\bigg) , then\: \dfrac{dy}{d\alpha} \: at\: \alpha=\dfrac{5\pi}{6}\: is$

Answer 1

Answer:

The Answer is -4.

Kindly refer to the attachment for the calculations.

Hope the answer is correct.

Answer 2

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

$\sf if \: y(\alpha)=\sqrt{2\bigg(\dfrac{tan\alpha+cot\alpha}{1+tan^2\alpha}\bigg) +\dfrac{1}{sin^2\alpha} } , \\ \sf \: \alpha \in \bigg(\dfrac{3\pi}{4},\pi\bigg) , then\: \dfrac{dy}{d\alpha} \: at\: \alpha=\dfrac{5\pi}{6}\: is \: \: \: \: \:$

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

$\rm :\longmapsto\:y = \sqrt{2\bigg(\dfrac{tan\alpha+cot\alpha}{1+tan^2\alpha}\bigg) +\dfrac{1}{sin^2\alpha} }$

$\rm :\longmapsto\:y = \sqrt{2\bigg(\dfrac{tan\alpha+\dfrac{1}{tan \alpha } }{ \: \: 1+tan^2\alpha \: \: }\bigg) +\dfrac{1}{sin^2\alpha} }$

$\rm :\longmapsto\:y = \sqrt{2\bigg(\dfrac{\dfrac{ {tan}^{2} \alpha + 1}{tan \alpha } }{ \: \: 1+tan^2\alpha \: \: }\bigg) +\dfrac{1}{sin^2\alpha} }$

$\rm :\longmapsto\:y = \sqrt{\bigg(\dfrac{2}{tan \alpha }{}\bigg) +\dfrac{1}{sin^2\alpha} }$

$\rm :\longmapsto\:y = \sqrt{2cot \alpha + {cosec}^{2} \alpha }$

$\: \: \: \because \: \boxed{ \pink{\bf{\dfrac{1}{tanx} = cotx \: and \: \dfrac{1}{sinx} = cosecx}}}$

$\rm :\longmapsto\:y = \sqrt{2cot \alpha + 1 + {cot}^{2} \alpha }$

$\rm :\longmapsto\:y = \sqrt{(cot \alpha + 1)^{2} }$

$\: \: \: \: \: \: \: \: \: \: \: \because \: \boxed{ \pink{\bf{ {x}^{2} + {y}^{2} + 2xy = {(x + y)}^{2}}}}$

$\rm :\implies\:y = |1 + cot \alpha |$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \because \: \boxed{ \pink{\bf{ \sqrt{ {x}^{2}} = |x| }}}$

$\red{\sf\:As\:\alpha \in \:\bigg(\dfrac{3\pi}{4},\pi \bigg) \: so \: cot \alpha < - 1 \implies \: 1 + cot \alpha < 0}$

$\bf\implies \:y = - (1 + cot \alpha )$

Now, on differentiating both sides, we get

$\rm :\longmapsto\:\dfrac{d}{d \alpha}y = - \dfrac{d}{d \alpha}(1 + cot \alpha )$

$\rm :\longmapsto\:\dfrac{dy}{d \alpha} = - ( - {cosec}^{2} \alpha)$

$\: \: \: \: \: \: \: \: \: \: \: \because \: \boxed{ \pink{\bf{\dfrac{d}{dx}cotx = - {cosec}^{2}x}}}$

$\rm :\longmapsto\:\dfrac{dy}{d \alpha} ={cosec}^{2} \alpha$

$\rm :\longmapsto\:\dfrac{dy}{d \alpha} \bigg(\: at \: \alpha =\dfrac{5\pi}{6} \bigg) ={cosec}^{2}\dfrac{5\pi}{6}$

$\rm :\longmapsto\:\dfrac{dy}{d \alpha}\bigg( \: at \: \alpha =\dfrac{5\pi}{6}\bigg)={cosec}^{2}(\pi - \dfrac{\pi}{6})$

$\rm :\longmapsto\:\dfrac{dy}{d \alpha} \bigg(\: at \: \alpha =\dfrac{5\pi}{6} \bigg) ={cosec}^{2}\dfrac{\pi}{6}$

$\rm :\longmapsto\:\dfrac{dy}{d \alpha} \: at \: \bigg(\alpha =\dfrac{5\pi}{6} \bigg) = {(2)}^{2} = 4$

Hence,

$\bf if \: y(\alpha)=\sqrt{2\bigg(\dfrac{tan\alpha+cot\alpha}{1+tan^2\alpha}\bigg) +\dfrac{1}{sin^2\alpha} } , \\ \bf \: \alpha \in \bigg(\dfrac{3\pi}{4},\pi\bigg) , then\: \dfrac{dy}{d\alpha} \: at\: \alpha=\dfrac{5\pi}{6}\: is \:4 \: \: \: \:$

Additional Information :-

$\boxed{ \red{\bf{\dfrac{d}{dx}tanx = {sec}^{2}x}}}$

$\boxed{ \red{\bf{\dfrac{d}{dx}cotx = - {cosec}^{2}x}}}$

$\boxed{ \red{\bf{\dfrac{d}{dx}cosx = - sinx}}}$

$\boxed{ \red{\bf{\dfrac{d}{dx}sinx = cosx}}}$

$\boxed{ \red{\bf{\dfrac{d}{dx}cosecx = - cosecxcotx}}}$

$\boxed{ \red{\bf{\dfrac{d}{dx}secx = secxtanx}}}$