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 $\alpha \in \left(\frac{3\mathrm{\pi }}{4},\mathrm{\pi }\right)$ then $\frac{dy}{d\alpha }$ at $\alpha =\frac{5\mathrm{\pi }}{6}$ is

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

Answer 1

The correct option is C

$4$

Explanation for the correct option:

Step 1. Given Information

The given function is $y\left(\alpha \right)=\sqrt{2\left(\frac{\tan \alpha +cot\alpha }{1+{\tan }^2\alpha }\right)+\frac{1}{{\sin }^2\alpha }}$where $\alpha \in \left(\frac{3\mathrm{\pi }}{4},\mathrm{\pi }\right)$

Step 2. Simplify the given equation.

Use $\tan \alpha =\frac{\sin \alpha }{\cos \alpha }$ and $\cot \alpha =\frac{\cos \alpha }{\sin \alpha }$ the function can be modified as:

$\begin{array}{rcl}y\left(\alpha \right)& =& \sqrt{2\left(\frac{\tan \alpha +cot\alpha }{1+{\tan }^2\alpha }\right)+\frac{1}{{\sin }^2\alpha }}\\ & =& \sqrt{2\left(\frac{{\displaystyle \frac{\sin \alpha }{\cos \alpha }}+{\displaystyle \frac{\cos \alpha }{\sin \alpha }}}{1+{\displaystyle \frac{{\sin }^2\alpha }{{\cos }^2\alpha }}}\right)+\frac{1}{{\sin }^2\alpha }}\\ & =& \sqrt{2\left(\frac{{\displaystyle \frac{{\sin }^2\alpha +{\cos }^2\alpha }{\sin \alpha \cos \alpha }}}{{\displaystyle \frac{{\cos }^2\alpha +{\sin }^2\alpha }{{\cos }^2\alpha }}}\right)+cose{c}^2\alpha }\\ & =& \sqrt{2\left(\frac{\cos \alpha }{\sin \alpha }\right)+1+{\cot }^2\alpha }\\ & =& \sqrt{2\cot \alpha +1+{\cot }^2\alpha }\\ & =& \sqrt{{\left(1+\cot \alpha \right)}^2}\\ & =& -1-\cot \alpha \end{array}$

Step 3. Find the value of $\frac{dy}{d\alpha }$ at $\alpha =\frac{5\mathrm{\pi }}{6}$.

Differentiate the function $y\left(\alpha \right)=-1-cot\alpha$ with respect to $x$

$\begin{array}{rcl}\frac{dy}{d\alpha }& =& \frac{d}{d\alpha }(-1-cot\alpha )\\ & =& 0-(-{\csc }^2\alpha )\\ & =& {\csc }^2\alpha \end{array}$

So at $\alpha =\frac{5\mathrm{\pi }}{6}$ the value of $\frac{dy}{d\alpha }$ is given as:

$\begin{array}{rcl}\frac{dy}{d\alpha }& =& {\csc }^2\left(\frac{5\mathrm{\pi }}{6}\right)\\ & =& 2^2\\ & =& 4\end{array}$

Hence, the correct option is (C).