Answer
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Hint:Firstly we will try to reduce $f\left( x \right)$ into a simpler form and then we will consider the equation of the normal to $f\left( x \right)$ at the given point $x = \dfrac{\pi }{6}$ and get the point required.
Complete step-by-step answer:
$f\left( x \right)$$ = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 + \sin x}}{{1 - \sin x}}} } \right)$ where $x \in \left( {0,\dfrac{\pi }{2}} \right)$$ - - - - - - (1)$
Consider $y = $$f(x)$$ = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 + \sin x}}{{1 - \sin x}}} } \right)$
Using $1 = {\sin ^2}x + {\cos ^2}x$ and $\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ we can write
$f(x) = {\tan ^{ - 1}}\sqrt {\dfrac{{{{\left( {\sin \dfrac{x}{2} + \cos \dfrac{x}{2}} \right)}^2}}}{{{{\left( {\sin \dfrac{x}{2} - \cos \dfrac{x}{2}} \right)}^2}}}} - - - - - - (2)$
$f(x) = {\tan ^{ - 1}}\sqrt {\dfrac{{{{\sin }^2}\dfrac{x}{2} + {{\cos }^2}\dfrac{x}{2} + 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}{{{{\sin }^2}\dfrac{x}{2} + {{\cos }^2}\dfrac{x}{2} - 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}} $
Using ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ where $a = \sin \dfrac{x}{2},b = \cos \dfrac{x}{2}$
Now it is given that $x \in \left( {0,\dfrac{\pi }{2}} \right)$
$0 < x < \dfrac{\pi }{2}$
$0 < \dfrac{x}{2} < \dfrac{\pi }{4}$
Therefore for $\dfrac{x}{2} \in \left( {0,\dfrac{\pi }{4}} \right)$, we know that,
$\cos \dfrac{x}{2} > \sin \dfrac{x}{2}$
$\cos \dfrac{x}{2} - \sin \dfrac{x}{2} > 0- - - - - - (3)$
Now by Substituting the value of equation 3 in equation 2 we will get.
$f\left( x \right) = {\tan ^{ - 1}}\left( {\dfrac{{\sin \dfrac{x}{2} + \cos \dfrac{x}{2}}}{{\cos \dfrac{x}{2} - \sin \dfrac{x}{2}}}} \right)$
$f\left( x \right) = {\tan ^{ - 1}}\left( {\dfrac{{(\tan \dfrac{x}{2} + 1)\cos \dfrac{x}{2}}}{{\cos \dfrac{x}{2}(1 - \tan \dfrac{x}{2})}}} \right)$
$f\left( x \right) = {\tan ^{ - 1}}\left( {\dfrac{{1 + \tan \dfrac{x}{2}}}{{1 - \tan \dfrac{x}{2}.1}}} \right)$
Using $\tan \dfrac{\pi }{4} = 1$
$f\left( x \right) = {\tan ^{ - 1}}\left( {\dfrac{{\tan \dfrac{\pi }{4} + \tan \dfrac{x}{2}}}{{\tan \dfrac{\pi }{4} - \tan \dfrac{x}{2}.\tan \dfrac{\pi }{4}}}} \right)$
${\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - a.b}}} \right) = {\tan ^{ - 1}}a + {\tan ^{ - 1}}b$
$f\left( x \right) = {\tan ^{ - 1}}\left( {\tan \dfrac{\pi }{4}} \right) + {\tan ^{ - 1}}\left( {\tan \dfrac{x}{2}} \right)$
Now we use
$a = {\tan ^{ - 1}}\left( {\tan a} \right)$
$y = f\left( x \right) = \left( {\dfrac{\pi }{4}} \right) + \left( {\dfrac{x}{2}} \right) - - - - - (4)$
Now take derivative of (4) with respect to $x$,
$f'(x) = 0 + \dfrac{1}{2} = \dfrac{1}{2}$
So slope of the $y = f\left( x \right) = \dfrac{1}{2}$
Slope of line normal to $f(x) = - 2$
So according to the question,
We need to find the equation of the normal to $y = f\left( x \right)$ at $x = \dfrac{\pi }{6}$
Firstly we need to put $x = \dfrac{\pi }{6}$ in equation (4) and get the value of $y$
$y = f\left( {\dfrac{\pi }{6}} \right) = \left( {\dfrac{\pi }{4}} \right) + \left( {\dfrac{\pi }{{2.6}}} \right) = \dfrac{\pi }{3}$
$y = \dfrac{\pi }{3}$
So normal to $y = f(x)$ is at the point $\left( {\dfrac{\pi }{6},\dfrac{\pi }{3}} \right)$
So equation of the normal will be:
$(y - \dfrac{\pi }{3}) = - 2(y - \dfrac{\pi }{6}) - - - - - - (5)$
Now we have to check the options by putting them in this equation:
Taking $(0,0)$
$(0 - \dfrac{\pi }{3}) = - 2(0 - \dfrac{\pi }{6})$
$ - \dfrac{\pi }{3} \ne \dfrac{\pi }{3}$
So it is not satisfying.
Now taking $(0,\dfrac{{2\pi }}{3})$
$(\dfrac{{2\pi }}{3} - \dfrac{\pi }{3}) = - 2(0 - \dfrac{\pi }{6})$
$\dfrac{\pi }{3} = \dfrac{\pi }{3}$
Hence $(0,\dfrac{{2\pi }}{3})$ is the correct option.
So, the correct answer is “Option B”.
Note:We need to know that if the slope of the line is $a$, then the slope of its normal will be $\dfrac{1}{a}$. Also that equation of the line passing through $(l,m)$ and slope $a$ is $(y - m) = a(x - l)$.
Complete step-by-step answer:
$f\left( x \right)$$ = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 + \sin x}}{{1 - \sin x}}} } \right)$ where $x \in \left( {0,\dfrac{\pi }{2}} \right)$$ - - - - - - (1)$
Consider $y = $$f(x)$$ = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 + \sin x}}{{1 - \sin x}}} } \right)$
Using $1 = {\sin ^2}x + {\cos ^2}x$ and $\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ we can write
$f(x) = {\tan ^{ - 1}}\sqrt {\dfrac{{{{\left( {\sin \dfrac{x}{2} + \cos \dfrac{x}{2}} \right)}^2}}}{{{{\left( {\sin \dfrac{x}{2} - \cos \dfrac{x}{2}} \right)}^2}}}} - - - - - - (2)$
$f(x) = {\tan ^{ - 1}}\sqrt {\dfrac{{{{\sin }^2}\dfrac{x}{2} + {{\cos }^2}\dfrac{x}{2} + 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}{{{{\sin }^2}\dfrac{x}{2} + {{\cos }^2}\dfrac{x}{2} - 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}} $
Using ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ where $a = \sin \dfrac{x}{2},b = \cos \dfrac{x}{2}$
Now it is given that $x \in \left( {0,\dfrac{\pi }{2}} \right)$
$0 < x < \dfrac{\pi }{2}$
$0 < \dfrac{x}{2} < \dfrac{\pi }{4}$
Therefore for $\dfrac{x}{2} \in \left( {0,\dfrac{\pi }{4}} \right)$, we know that,
$\cos \dfrac{x}{2} > \sin \dfrac{x}{2}$
$\cos \dfrac{x}{2} - \sin \dfrac{x}{2} > 0- - - - - - (3)$
Now by Substituting the value of equation 3 in equation 2 we will get.
$f\left( x \right) = {\tan ^{ - 1}}\left( {\dfrac{{\sin \dfrac{x}{2} + \cos \dfrac{x}{2}}}{{\cos \dfrac{x}{2} - \sin \dfrac{x}{2}}}} \right)$
$f\left( x \right) = {\tan ^{ - 1}}\left( {\dfrac{{(\tan \dfrac{x}{2} + 1)\cos \dfrac{x}{2}}}{{\cos \dfrac{x}{2}(1 - \tan \dfrac{x}{2})}}} \right)$
$f\left( x \right) = {\tan ^{ - 1}}\left( {\dfrac{{1 + \tan \dfrac{x}{2}}}{{1 - \tan \dfrac{x}{2}.1}}} \right)$
Using $\tan \dfrac{\pi }{4} = 1$
$f\left( x \right) = {\tan ^{ - 1}}\left( {\dfrac{{\tan \dfrac{\pi }{4} + \tan \dfrac{x}{2}}}{{\tan \dfrac{\pi }{4} - \tan \dfrac{x}{2}.\tan \dfrac{\pi }{4}}}} \right)$
${\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - a.b}}} \right) = {\tan ^{ - 1}}a + {\tan ^{ - 1}}b$
$f\left( x \right) = {\tan ^{ - 1}}\left( {\tan \dfrac{\pi }{4}} \right) + {\tan ^{ - 1}}\left( {\tan \dfrac{x}{2}} \right)$
Now we use
$a = {\tan ^{ - 1}}\left( {\tan a} \right)$
$y = f\left( x \right) = \left( {\dfrac{\pi }{4}} \right) + \left( {\dfrac{x}{2}} \right) - - - - - (4)$
Now take derivative of (4) with respect to $x$,
$f'(x) = 0 + \dfrac{1}{2} = \dfrac{1}{2}$
So slope of the $y = f\left( x \right) = \dfrac{1}{2}$
Slope of line normal to $f(x) = - 2$
So according to the question,
We need to find the equation of the normal to $y = f\left( x \right)$ at $x = \dfrac{\pi }{6}$
Firstly we need to put $x = \dfrac{\pi }{6}$ in equation (4) and get the value of $y$
$y = f\left( {\dfrac{\pi }{6}} \right) = \left( {\dfrac{\pi }{4}} \right) + \left( {\dfrac{\pi }{{2.6}}} \right) = \dfrac{\pi }{3}$
$y = \dfrac{\pi }{3}$
So normal to $y = f(x)$ is at the point $\left( {\dfrac{\pi }{6},\dfrac{\pi }{3}} \right)$
So equation of the normal will be:
$(y - \dfrac{\pi }{3}) = - 2(y - \dfrac{\pi }{6}) - - - - - - (5)$
Now we have to check the options by putting them in this equation:
Taking $(0,0)$
$(0 - \dfrac{\pi }{3}) = - 2(0 - \dfrac{\pi }{6})$
$ - \dfrac{\pi }{3} \ne \dfrac{\pi }{3}$
So it is not satisfying.
Now taking $(0,\dfrac{{2\pi }}{3})$
$(\dfrac{{2\pi }}{3} - \dfrac{\pi }{3}) = - 2(0 - \dfrac{\pi }{6})$
$\dfrac{\pi }{3} = \dfrac{\pi }{3}$
Hence $(0,\dfrac{{2\pi }}{3})$ is the correct option.
So, the correct answer is “Option B”.
Note:We need to know that if the slope of the line is $a$, then the slope of its normal will be $\dfrac{1}{a}$. Also that equation of the line passing through $(l,m)$ and slope $a$ is $(y - m) = a(x - l)$.