Question

If $y=\log \left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$, then $\frac{dy}{dx}$ is equal to

• A. $\sec x$
• B. $\tan x$
• C. $\sec 2x$
• D. none of these

Answer 1

The correct option is D none of these

$y=\log \left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)----\left(1\right)$

Now, $1+\sin x={\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+2\sin x/2\cos x/2$

$=(\sin x/2+\cos x/2{)}^2$

Similarly, $1-\sin x=(\sin x/2-\cos x/2{)}^2$

$\therefore$From $\left(1\right)$, taking only positive values,

$yu=\log \left(\frac{\sin x/2+\cos x/2}{\sin x/2-\cos x/2}\right)$

or, $y=\log \left(\frac{\tan x/2+1}{\tan x/2-1}\right)$

or, $y=\log \left\{\tan (\frac{\pi }{4}+\frac{x}{2})\right\}$

$\therefore \frac{dy}{dx}=\frac{1}{\tan (\frac{\pi }{4}+\frac{x}{2})}.{\sec }^2(\frac{\pi }{4}+\frac{x}{2})\times \frac{1}{2}$

$=\frac{1}{2}{\sec }^2(\frac{\pi }{4}+\frac{x}{2})\cot (\frac{\pi }{4}+\frac{x}{2})$