Consider f(x) = tan^{-1}left(sqrt{frac{1 + si | vedclass

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(D) Given $f(x) = 	an^{-1}\left(\sqrt{\frac{1 + \sin x}{1 - \sin x}}ight)$.Using $1 + \sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$ and $1 - \s

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$(0, \frac{2\pi}{3})$

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(D) Given $f(x) = 	an^{-1}\left(\sqrt{\frac{1 + \sin x}{1 - \sin x}}ight)$.Using $1 + \sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$ and $1 - \sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$,we have:$f(x) = 	an^{-1}\left(\frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}ight) = 	an^{-1}\left(	an(\frac{\pi}{4} + \frac{x}{2})ight) = \frac{\pi}{4} + \frac{x}{2}$.At $x = \frac{\pi}{6}$,$f(\frac{\pi}{6}) = \frac{\pi}{4} + \frac{\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3}$.The derivative $f'(x) = \frac{1}{2}$.The slope of the tangent at $x = \frac{\pi}{6}$ is $m_t = \frac{1}{2}$.The slope of the normal is $m_n = -\frac{1}{m_t} = -2$.The equation of the normal at $(\frac{\pi}{6}, \frac{\pi}{3})$ is $y - \frac{\pi}{3} = -2(x - \frac{\pi}{6})$.$y - \frac{\pi}{3} = -2x + \frac{\pi}{3} \Rightarrow y = -2x + \frac{2\pi}{3}$.Checking the points,if $x = 0$,$y = \frac{2\pi}{3}$. Thus,it passes through $(0, \frac{2\pi}{3})$.

Consider $f(x) = \tan^{-1}\left(\sqrt{\frac{1 + \sin x}{1 - \sin x}}\right)$,where $x \in (0, \frac{\pi}{2})$. $A$ normal to $y = f(x)$ at $x = \frac{\pi}{6}$ also passes through the point:

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