Question

If $y={\tan }^{-1}\frac{\overline{1+\sin x}}{1-\sin x},\frac{\pi }{2}<x<\pi$, then $\frac{dy}{dx}$ equals

• A. $\frac{-1}{2}$
• B. $-1$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

Answer: (A)

$\frac{-1}{2}$

$y={\tan }^{-1}\frac{\overline{1-\cos (\frac{\pi }{2}+x)}}{1+\cos (\frac{\pi }{2}+x)}$

$={\tan }^{-1}\left|\tan (\frac{\pi }{4}+\frac{x}{2})\right|$ ...(i)

Now, $\frac{\pi }{2}<x<\pi$

$\therefore \frac{\pi }{4}<\frac{x}{2}<\frac{\pi }{2}$

or $\frac{\pi }{2}<\frac{\pi }{4}+\frac{x}{2}<\frac{3\pi }{4}$

$\therefore \left|\tan (\frac{\pi }{4}+\frac{}{})\right|=\tan (\frac{\pi }{4}+\frac{x}{2})$ (Q in II quadrant)

$=\tan \{\pi -(\frac{\pi }{4}+\frac{x}{2})\}$

from Eq. (i);

$y={\tan }^{-1}\tan \{\pi -(\frac{\pi }{4}+\frac{\pi }{2})\}$

$=\pi -(\frac{\pi }{4}+\frac{\pi }{2})$

$=\frac{3\pi }{4}-\frac{x}{2}$

$(\text{Q principle value of}{\tan }^{-1}\text{is}-\frac{\pi }{2}\text{to}\frac{\pi }{2})$

Therefore, $\frac{dy}{dx}=\frac{d}{dx}(\frac{3\pi }{4}-\frac{x}{2})=-\frac{1}{2}$