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Question: Consider\(f(x)\) \( = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 + \sin x}}{{1 - \sin x}}} } \right)\) w...

Considerf(x)f(x) =tan⁡−1(1+sin⁡x1−sin⁡x) = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 + \sin x}}{{1 - \sin x}}} } \right) wherex∈(0,π2) x \in \left( {0,\dfrac{\pi }{2}} \right) . A normal to y=f(x)y = f(x)at x=π6x = \dfrac{\pi }{6}also passes through the point:
A. (0,0)(0,0)
B. (0,2π3)\left( {0,\dfrac{{2\pi }}{3}} \right)
C. (π6,0)\left( {\dfrac{\pi }{6},0} \right)
D. (π4,0)\left( {\dfrac{\pi }{4},0} \right)

Explanation

Solution

Firstly we will try to reduce f(x)f\left( x \right) into a simpler form and then we will consider the equation of the normal to f(x)f\left( x \right) at the given point x=π6x = \dfrac{\pi }{6} and get the point required.

Complete step-by-step answer:
f(x)f\left( x \right) =tan⁡−1(1+sin⁡x1−sin⁡x) = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 + \sin x}}{{1 - \sin x}}} } \right) where x∈(0,π2)x \in \left( {0,\dfrac{\pi }{2}} \right) −−−−−−(1) - - - - - - (1)
Consider y=y = f(x)f(x) =tan⁡−1(1+sin⁡x1−sin⁡x) = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 + \sin x}}{{1 - \sin x}}} } \right)
Using 1=sin⁡2x+cos⁡2x1 = {\sin ^2}x + {\cos ^2}x and sin⁡x=2sin⁡x2cos⁡x2\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} we can write
f(x)=tan⁡−1(sin⁡x2+cos⁡x2)2(sin⁡x2−cos⁡x2)2−−−−−−(2)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⁡−1sin⁡2x2+cos⁡2x2+2sin⁡x2cos⁡x2sin⁡2x2+cos⁡2x2−2sin⁡x2cos⁡x2f(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 (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab where a=sin⁡x2,b=cos⁡x2a = \sin \dfrac{x}{2},b = \cos \dfrac{x}{2}

Now it is given that x∈(0,π2)x \in \left( {0,\dfrac{\pi }{2}} \right)
0<x<π20 < x < \dfrac{\pi }{2}
0<x2<π40 < \dfrac{x}{2} < \dfrac{\pi }{4}
Therefore for x2∈(0,π4)\dfrac{x}{2} \in \left( {0,\dfrac{\pi }{4}} \right), we know that,
cos⁡x2>sin⁡x2\cos \dfrac{x}{2} > \sin \dfrac{x}{2}
cos⁡x2−sin⁡x2>0−−−−−−(3)\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(x)=tan⁡−1(sin⁡x2+cos⁡x2cos⁡x2−sin⁡x2)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(x)=tan⁡−1((tan⁡x2+1)cos⁡x2cos⁡x2(1−tan⁡x2))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(x)=tan⁡−1(1+tan⁡x21−tan⁡x2.1)f\left( x \right) = {\tan ^{ - 1}}\left( {\dfrac{{1 + \tan \dfrac{x}{2}}}{{1 - \tan \dfrac{x}{2}.1}}} \right)
Using tan⁡π4=1\tan \dfrac{\pi }{4} = 1
f(x)=tan⁡−1(tan⁡π4+tan⁡x2tan⁡π4−tan⁡x2.tan⁡π4)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(a+b1−a.b)=tan⁡−1a+tan⁡−1b{\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - a.b}}} \right) = {\tan ^{ - 1}}a + {\tan ^{ - 1}}b
f(x)=tan⁡−1(tan⁡π4)+tan⁡−1(tan⁡x2)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(tan⁡a)a = {\tan ^{ - 1}}\left( {\tan a} \right)
y=f(x)=(π4)+(x2)−−−−−(4)y = f\left( x \right) = \left( {\dfrac{\pi }{4}} \right) + \left( {\dfrac{x}{2}} \right) - - - - - (4)
Now take derivative of (4) with respect to xx,
f′(x)=0+12=12f'(x) = 0 + \dfrac{1}{2} = \dfrac{1}{2}
So slope of the y=f(x)=12y = f\left( x \right) = \dfrac{1}{2}
Slope of line normal to f(x)=−2f(x) = - 2
So according to the question,
We need to find the equation of the normal to y=f(x)y = f\left( x \right) at x=π6x = \dfrac{\pi }{6}
Firstly we need to put x=π6x = \dfrac{\pi }{6} in equation (4) and get the value of yy
y=f(π6)=(π4)+(π2.6)=π3y = f\left( {\dfrac{\pi }{6}} \right) = \left( {\dfrac{\pi }{4}} \right) + \left( {\dfrac{\pi }{{2.6}}} \right) = \dfrac{\pi }{3}
y=π3y = \dfrac{\pi }{3}
So normal to y=f(x)y = f(x) is at the point (π6,π3)\left( {\dfrac{\pi }{6},\dfrac{\pi }{3}} \right)
So equation of the normal will be:
(y−π3)=−2(y−π6)−−−−−−(5)(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)(0,0)
(0−π3)=−2(0−π6)(0 - \dfrac{\pi }{3}) = - 2(0 - \dfrac{\pi }{6})
−π3≠π3- \dfrac{\pi }{3} \ne \dfrac{\pi }{3}
So it is not satisfying.
Now taking (0,2π3)(0,\dfrac{{2\pi }}{3})
(2π3−π3)=−2(0−π6)(\dfrac{{2\pi }}{3} - \dfrac{\pi }{3}) = - 2(0 - \dfrac{\pi }{6})
π3=π3\dfrac{\pi }{3} = \dfrac{\pi }{3}
Hence (0,2π3)(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 aa, then the slope of its normal will be 1a\dfrac{1}{a}. Also that equation of the line passing through (l,m)(l,m) and slope aa is (y−m)=a(x−l)(y - m) = a(x - l).