Question

If $y=\log \tan \left(\frac{\pi }{4}+\frac{x}{2}\right),$ then $\frac{dy}{dx}=$

• A. $\sec x$
• B. $\sin x$
• C. $cosecx$
• D. $\sec \frac{x}{2}$

Answer 1

Answer: (A)

Given, $y=\log \tan \left(\frac{\pi }{4}+\frac{x}{2}\right)$
$\Rightarrow \frac{dy}{dx}=\frac{1}{\tan \left(\frac{\pi }{4}+\frac{x}{2}\right)}\frac{1}{2}{\sec }^2\left(\frac{\pi }{4}+\frac{x}{2}\right)$
$=\frac{1}{2}\left[\frac{1}{\tan \left(\frac{\pi }{4}+\frac{x}{2}\right)}+\frac{{\tan }^2\left(\frac{\pi }{4}+\frac{x}{2}\right)}{\tan \left(\frac{\pi }{4}+\frac{x}{2}\right)}\right]=$
$=\frac{1}{2}\left[\frac{\cos \left(\frac{\pi }{4}+\frac{x}{2}\right)}{\sin \left(\frac{\pi }{4}+\frac{x}{2}\right)}+\frac{\sin \left(\frac{\pi }{4}+\frac{x}{2}\right)}{\cos \left(\frac{\pi }{4}+\frac{x}{2}\right)}\right]$
$\frac{dy}{dx}=\frac{1}{\sin \left(\frac{\pi }{2}+x\right)}\left[\because {\cos }^{2}x{\sin }^{2}x=1\right]$
$=\frac{1}{\cos x}=\sec x\left[\because \sin \left(\frac{\pi }{2}+\theta \right)=\cos \theta \right]$