MCQ
If $\mathrm{y}(\alpha)=\sqrt{2\left(\frac{\tan \alpha+\cot \alpha}{1+\tan ^{2} \alpha}\right)+\frac{1}{\sin ^{2} \alpha}}, \alpha \in\left(\frac{3 \pi}{4}, \pi\right)$ then $\frac{d y}{d \alpha}$ at $\alpha=\frac{5 \pi}{6}$ is
  • $4$
  • B
    $-\frac{1}{4}$
  • C
    $\frac{4}{3}$
  • D
    $-4$

Answer

Correct option: A.
$4$
a
$\mathrm{y}(\alpha)=\sqrt{2 \frac{(\tan \alpha+\cot \alpha)}{1+\tan ^{2} \alpha}+\frac{1}{\sin ^{2} \alpha}}, \alpha \in\left(\frac{3 \pi}{4}, \pi\right)$

$=\frac{|\sin \alpha+\cos \alpha|}{|\sin \alpha|}=\frac{-(\sin \alpha+\cos \alpha)}{\sin \alpha}$

$=-1-\cot \alpha$

$y^{\prime}(\alpha)=cosec ^{2} \alpha$

$y^{\prime}\left(\frac{5 \pi}{6}\right)=4$

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