Question

$\frac{d}{dx}\left[\log \sqrt{\frac{1-cosx}{1+cosx}}\right]$=

• A. $secx$
• B. $cscx$
• C. $cosec\frac{x}{2}$
• D. $sec\frac{x}{2}$

Answer 1

The correct option is B $cscx$

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

$\Rightarrow \log \sqrt{\frac{1-\cos x}{1+\cos x}}\sqrt{\frac{1-\cos x}{1-\cos x}}$

$\Rightarrow \log \left(\frac{1-\cos x}{\sin x}\right)\Rightarrow \log \left(\frac{2{\cos }^{2}x/2}{2\sin x/2\cos x/2}\right)$

$\Rightarrow \log \left(\cot x/2\right)$

$\Rightarrow \frac{d}{dx}\left(\log \left(\cot \left(\frac{x}{2}\right)\right)\right)$

$=\frac{-1}{\cot x/2}{\csc }^{2}x/2\left(\frac{1}{2}\right)$

$=-\frac{1}{2}\frac{\cos x/2}{\sin x/2}\frac{1}{{\sin }^{2}x/2}$

$=-\frac{1}{2}\tan \left(\frac{x}{2}\right)\csc \left(\frac{x}{2}\right)$