Question

If $y=ta{n}^{-1}\sqrt{\frac{1-sinx}{1+sinx}}$ then the value of $\frac{dy}{dx}$ at $x=\frac{\pi }{6}$ is

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

Answer 1

The correct option is A

$-\frac{1}{2}$

$\frac{1-sinx}{1+sinx}=\frac{{(cos\frac{x}{2}-sin\frac{x}{2})}^2}{{(cos\frac{x}{2}+sin\frac{x}{2})}^2}$
$cos\frac{x}{2}>sin\frac{x}{2}$
$\therefore ta{n}^{-1}\left(\sqrt{\frac{1-sinx}{1+sinx}}\right)=ta{n}^{-1}\left(\frac{cos\frac{x}{2}-sin\frac{x}{2}}{cos\frac{x}{2}+sin\frac{x}{2}}\right)=ta{n}^{-1}\left(\frac{1-tan\frac{x}{2}}{1+tan\frac{x}{2}}\right)$
$=ta{n}^{-1}\left(tan(\frac{\pi }{4}-\frac{x}{2})\right)$
$=\frac{\pi }{4}-\frac{x}{2}$ = y
$\left(\frac{dy}{dx}\right)$ = $\left(\frac{-1}{2}\right)$