Question

If y = $ta{n}^{-1}\sqrt{\left(\frac{1+sinx}{1-sinx}\right)},\frac{\pi }{2}<x<\pi$, then $\frac{dy}{dx}$ equals

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

Answer 1

The correct option is A $\frac{-1}{2}$

$y=ta{n}^{-1}\sqrt{\left(\frac{1-cos(\frac{\pi }{2}+x)}{1+cos(\frac{\pi }{2}+x)}\right)}=ta{n}^{-1}\left|tan(\frac{\pi }{4}+\frac{x}{2})\right|\cdots \left)\right(i)$
$Now,\frac{\pi }{2}<x<\pi$
$\therefore \frac{\pi }{4}<\frac{x}{2}<\frac{\pi }{2}$
$or\frac{\pi }{2}<\frac{\pi }{4}+\frac{x}{2}<\frac{3\pi }{4}$
$\therefore \left|tan(\frac{\pi }{4}+\frac{x}{2})\right|=-tan(\frac{\pi }{4}+\frac{x}{2})(\therefore insecondquadrant)$
$=tan\{\pi -(\frac{\pi }{4}+\frac{x}{2})\}$
$FromEq.\left(i\right),$
$y=ta{n}^{-1}tan\{\pi -(\frac{\pi }{4}+\frac{x}{2})\}$
$=\pi -(\frac{\pi }{4}+\frac{x}{2})$
$=\frac{3\pi }{4}-\frac{x}{2}$
$(\because principalvalueofta{n}^{-1}xin-\frac{\pi }{2}to\frac{\pi }{2})$
$\therefore \frac{dy}{dx}=-\frac{1}{2}$