Question

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

• A. $\sec \: x$
• B. $\sin\: x$
• C. $cosec \: x$
• 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] $
$= \frac{1}{\cos x} =\sec x \left[ \because \:\: \sin \left( \frac{\pi }{2} + \theta\right) =\cos \theta\right]$