Question

If $y={\tan }^{-1}\sqrt{\frac{(1-\sin x)}{(1+\sin x)}}$,then the value of $\frac{dy}{dx}$at $x=\frac{\pi }{6}$is

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

Answer 1

The correct option is B

$\frac{-1}{2}$

Explanation for the correct option:

$$\begin{gathered} y={\tan }^{-1}\sqrt{\frac{\left(\begin{array}{cc}1-& \sin x\end{array}\right)}{\left(\begin{array}{cc}1+& \sin x\end{array}\right)}} \\ ={\tan }^{-1}\sqrt{\frac{1-\cos \left(\begin{array}{cc}{\displaystyle \frac{\pi }{2}}-& x\end{array}\right)}{\begin{array}{cc}1+& \cos \left(\begin{array}{cc}{\displaystyle \frac{\pi }{2}}-& x\end{array}\right)\end{array}}} \\ ={\tan }^{-1}\sqrt{\frac{2{\sin }^2\left(\begin{array}{cc}{\displaystyle \frac{\pi }{4}}-& {\displaystyle \frac{x}{2}}\end{array}\right)}{2{\cos }^2\left(\begin{array}{cc}{\displaystyle \frac{\pi }{4}}-& {\displaystyle \frac{x}{2}}\end{array}\right)}} \\ ={\tan }^{-1}\sqrt{{\tan }^2\left(\begin{array}{cc}\frac{\pi }{4}-& \frac{x}{2}\end{array}\right)} \\ ={\tan }^{-1}\tan \left(\begin{array}{cc}\frac{\pi }{4}-& \frac{x}{2}\end{array}\right) \\ =\frac{\pi }{4}-\frac{x}{2} \\ \Rightarrow y=\frac{\pi }{4}-\frac{x}{2} \end{gathered}$$

Now, differentiate with respect to $x$

$\frac{dy}{dx}=\frac{-1}{2}$

So, $\frac{dy}{dx}$at $x=\frac{\pi }{6}$ is $\frac{-1}{2}$.

Hence, Option (B) is correct answer.