Application Of Derivatives Ques 25
25. 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: 25.(b)
Solution:
Formula:
Equation of tangent and normal :
- 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 $0, \frac{2 \pi}{3}$.
BETA
