Question:
Mark $(\sqrt{ })$ against the correct answer in the following:
If $y=\sqrt{\frac{1+\tan x}{1-\tan x}}$ then $\frac{d y}{d x}=$ ?
A. $\frac{1}{2} \sec ^{2} x \cdot \tan \left(x+\frac{\pi}{4}\right)$
B. $\frac{\sec ^{2}\left(x+\frac{\pi}{4}\right)}{2 \sqrt{\tan \left(x+\frac{\pi}{4}\right)}}$
C. $\frac{\sec ^{2}\left(\frac{x}{4}\right)}{\sqrt{\tan \left(x+\frac{\pi}{4}\right)}}$
D. none of these
Solution:
