Application of Derivatives 1 Question 9
####9. Consider $f(x)=\tan ^{-1} \sqrt{\frac{1+\sin x}{1-\sin x}}, x \in 0, \frac{\pi}{2}$.
A normal to $y=f(x)$ at $x=\frac{\pi}{6}$ also passes through the point
(a) $(0,0)$
(b) $0, \frac{2 \pi}{3}$
(c) $\frac{\pi}{6}, 0$
(d) $\frac{\pi}{4}, 0$
(2016 Main)
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Answer:
Correct Answer: 9. $H=\varphi, V={1,1}$
Solution:
- We have, $f(x)=\tan ^{-1} \sqrt{\frac{1+\sin x}{1-\sin x}}, x \in 0, \frac{\pi}{2}$
$ \begin{aligned} \Rightarrow \quad f(x) & =\tan ^{-1} \sqrt{\frac{\cos \frac{x}{2}+\sin \frac{x^{2}}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}} \\ & =\tan ^{-1} \frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}} \\ & \because \cos \frac{x}{2}>\sin \frac{x}{2} \text { for } 0<\frac{x}{2}<\frac{\pi}{4} \\ & =\tan ^{-1} \frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}} \\ & =\tan ^{-1} \tan \frac{\pi}{4}+\frac{x}{2}=\frac{\pi}{4}+\frac{x}{2} \\ f^{\prime}(x) & =\frac{1}{2} \Rightarrow f^{\prime} \frac{\pi}{6}=\frac{1}{2} \end{aligned} $
Now, equation of normal at $x=\frac{\pi}{6}$ is given by
$ \begin{aligned} & y-f \frac{\pi}{6}=-2 \quad x-\frac{\pi}{6} \\ & \Rightarrow \quad y-\frac{\pi}{3}=-2 \quad x-\frac{\pi}{6} \quad \because f \quad \frac{\pi}{6}=\frac{\pi}{4}+\frac{\pi}{12}=\frac{4 \pi}{12}=\frac{\pi}{3} \end{aligned} $
which passes through $0, \frac{2 \pi}{3}$.
