Question

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

• A. $-\frac{1}{4}$
• B. $\frac{4}{3}$
• C. $4$
• D. $-4$

Answer 1

The correct option is C $4$

$y\left(\alpha \right)=\sqrt{2\left(\frac{\tan \alpha +\cot \alpha }{1+{\tan }^2\alpha }\right)+\frac{1}{{\sin }^2\alpha }}$

$\Rightarrow y\left(\alpha \right)=\sqrt{\frac{2}{\sin \alpha \cos \alpha \times \frac{1}{{\cos }^2\alpha }}+\frac{1}{{\sin }^2\alpha }}$

$\Rightarrow y\left(\alpha \right)=\sqrt{2\cot \alpha +{\text{cosec}}^2\alpha }\Rightarrow y\left(\alpha \right)=\sqrt{(1+\cot \alpha {)}^2}\Rightarrow y\left(\alpha \right)=-1-\cot \alpha [\because \alpha \in (\frac{3\pi }{4},\pi )]\frac{dy}{d\alpha }=0+{\text{cosec}}^2\alpha \Rightarrow \frac{dy}{d\alpha }{|}_{x=\frac{5\pi }{6}}={\text{cosec}}^2\frac{5\pi }{6}\Rightarrow \frac{dy}{d\alpha }{|}_{x=\frac{5\pi }{6}}=4$