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Question:

If $y(\alpha)=\sqrt{2\left(\frac{\tan \alpha+\cot \alpha}{1+\tan ^{2} \alpha}\right)+\frac{1}{\sin ^{2} \alpha}}, \alpha \in\left(\frac{3 \pi}{4}, \pi\right)$,

then $\frac{d y}{d \alpha}$ at $\alpha=\frac{5 \pi}{6}$ is:

  1. (1) 4

  2. (2) $\frac{4}{3}$

  3. (3) $-4$

  4. (4) $-\frac{1}{4}$


Correct Option: 1

Solution:

$y(\alpha)=\sqrt{\frac{\frac{2 \sin \alpha}{\cos \alpha}+\frac{\cos \alpha}{\sin \alpha}}{\sec ^{2} \alpha}}=\sqrt{\frac{2 \cos ^{2} \alpha}{\sin \alpha \cos \alpha}+\frac{1}{\sin ^{2} \alpha}}$

$=\sqrt{2 \cot \alpha+\operatorname{cosec}^{2} \alpha}=\sqrt{2 \cot \alpha+1+\cot ^{2} \alpha}$

$=|1+\cot \alpha|=-1-\cot \alpha \quad\left[\because \alpha \in\left(\frac{3 \pi}{4}, \pi\right)\right]$

$\frac{d y}{d \alpha}=\operatorname{cosec}^{2} \alpha \Rightarrow\left(\frac{d y}{d \alpha}\right)_{\alpha=\frac{5 \pi}{6}}=4$

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