Question

$\frac{d}{dx}\left[\log \sqrt{\frac{1-\cos \left(x\right)}{1+\cos \left(x\right)}}\right]=$

• A. $\cos ecx$
• B. $secx$
• C. $\cos ecx/2$
• D. $secx/2$

Answer 1

The correct option is A

$\cos ecx$

Explanation for the correct option:

Step 1. Find the differentiation of given function:

Given, $\frac{d}{dx}\left[\log \sqrt{\frac{1-\cos \left(x\right)}{1+\cos \left(x\right)}}\right]$

Let,

$\begin{array}{rcl}y& =& \left[\log \sqrt{\frac{1-\cos \left(x\right)}{1+\cos \left(x\right)}}\right]\\ & =& \left[\log \sqrt{\frac{1-\left(1-{\sin }^{2}x/2\right)}{1+{\cos }^{2}x/2-1}}\right]\\ & =& \log \left(\frac{\sqrt{{\sin }^2\left(x/2\right)}}{\sqrt{{\cos }^2(x/2)}}\right)\\ & =& \log \left(\sqrt{{\tan }^2\left(x/2\right)}\right)\\ & & \end{array}$

$\Rightarrow$$\begin{array}{rcl}y& =& \log \left(\tan \left(x/2\right)\right)\end{array}$

Step 2. Differentiate with respect to $\text{'}x\text{'}$

$\frac{dy}{dx}=\frac{1}{\tan \left(\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}\times \frac{d\left(se{c}^2{\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}{dx}\times \frac{d\left({\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}{dx}$

$\begin{array}{rcl}& =& \frac{1}{\tan x/2}\times se{c}^{2}x/2\times \frac{1}{2}\\ & =& \frac{\cos \left({\displaystyle \frac{x}{2}}\right)}{\sin \left({\displaystyle \frac{x}{2}}\right)}\times \frac{1}{{\cos }^2\left({\displaystyle \frac{x}{2}}\right)}\times \frac{1}{2}\\ & =& \frac{1}{2.\sin x/2.\cos x/2}\\ & =& \frac{1}{\sin x}\\ & & \end{array}$

$\Rightarrow$$\begin{array}{rcl}\frac{dy}{dx}& =& cosecx\end{array}$

$\therefore \begin{array}{rcl}& & \frac{\mathbf{d}\mathbf{\left[}\mathbf{log}\sqrt{\frac{\mathbf{1}\mathbf{-}\mathbf{cos}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{1}\mathbf{+}\mathbf{cos}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}}\mathbf{\right]}}{\mathbf{d}\mathbf{x}}\end{array}=\mathit{c}\mathit{o}\mathit{s}\mathit{e}\mathit{c}\mathbf{}\mathit{x}$

Hence, option(A) is correct.