Question

For $0<x<2$, $\frac{d({\tan }^{-1}\sqrt{\left({\displaystyle \frac{1+\cos {\displaystyle \frac{1}{2}}}{1-\cos \frac{1}{2}}}\right)}}{dx}$is equal to

• A. $\frac{-1}{4}$
• B. $\frac{1}{4}$
• C. $\frac{-1}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A

$\frac{-1}{4}$

Explanation for the correct option :

Step1. Simplifying the given equation

Let, $y={\tan }^{-1}\sqrt{\frac{1+\cos {\displaystyle \frac{x}{2}}}{1-\cos {\displaystyle \frac{x}{2}}}}...\left(i\right)$ $$\begin{gathered} \mathbf{(}\mathbf{1}\mathbf{+}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{x}\mathbf{)}\mathbf{=}\mathbf{2}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{x} \\ \mathbf{(}\mathbf{1}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{x}\mathbf{)}\mathbf{=}\mathbf{2}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{x} \end{gathered}$$

Now using the above cos formula.

$$\begin{gathered} 1+\cos \frac{x}{2}=2{\cos }^2\frac{x}{4}\& \\ 1-\cos \frac{x}{2}=2{\sin }^2\frac{x}{4} \end{gathered}$$

Insert this in equation $\left(i\right)$

Now,

$\begin{array}{rcl}y& =& {\tan }^{-1}\sqrt{\frac{2{\cos }^2\left({\displaystyle \frac{x}{4}}\right)}{2{\sin }^2\left({\displaystyle \frac{x}{4}}\right)}}\\ & =& {\tan }^{-1}\sqrt{co{t}^2\left(\frac{x}{4}\right)}\\ & =& {\tan }^{-1}\left[cot\right(\frac{\pi }{2}-\frac{x}{4})\\ & =& {\tan }^{-1}\left(\tan \right(\frac{\pi }{2}-\frac{x}{4}))\\ & =& \frac{\pi }{2}-\frac{x}{4}\\ & =& \frac{-x}{4}\end{array}$

Step3. Finding differentiation

Therefore, $\frac{dy}{dx}=\frac{-1}{4}$

Hence, correct option is $\mathbf{\left(}\mathit{A}\mathbf{\right)}$