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Q. If $y\left(\alpha\right) = \sqrt{2\left(\frac{tan\alpha+cot\alpha}{1+tan^{2}\alpha}\right)+\frac{1}{sin^{2}\alpha}}, \alpha \epsilon \left(\frac{3\pi}{4},\pi\right),$ then $\frac{dy}{d\alpha }$ at $\alpha = \frac{5\pi}{6}$ is :

JEE MainJEE Main 2020Continuity and Differentiability

Solution:

$y\left(\alpha\right) = \sqrt{2\left(\frac{tan\alpha+cot\alpha}{1+tan^{2}\alpha}\right)+\frac{1}{sin^{2}\alpha}}, \alpha \epsilon \left(\frac{3\pi}{4},\pi\right)$
$ =\frac{\left|sin\,\alpha + cos\,\alpha\right|}{\left|sin\,\alpha \right|} = \frac{-\left(sin\,\alpha +cos\,\alpha \right)}{sin\,\alpha }$
$= - 1 -cos\,\alpha $
$y' \left(\alpha\right) = cosec^{2}\alpha$
$y'\left(\frac{5\pi}{6}\right) = 4$