Question

Consider $f(x) = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 + \sin x}}{{1 - \sin x}}} } \right),x \in \left( {0,\dfrac{\pi }{2}} \right)$, a normal to $y = f(x)$ at $x = \dfrac{\pi }{6}$ also passes through the point:
A. $(0,0)$
B. $(0,\dfrac{{2\pi }}{3})$
C. $(\dfrac{\pi }{6},0)$
D. $(\dfrac{\pi }{4},0)$

Answer 1

We use the given function and then substitute the value of ‘x’ to obtain the value of function at that point. Solve the given function using the concept of complementary angles. Use the formulas of $1 - \cos 2\theta = 2{\sin ^2}\theta$ and $1 + \cos 2\theta = 2{\cos ^2}\theta$ to convert the fraction under the square root in terms of cotangent. Use the concept of complementary angles again to convert cotangent function into tangent function so as to cancel inverse of tangent given. Calculate slope of tangent by differentiating the function obtained. Use the formula of slope of normal to calculate slope of normal. Form equation of slope using point obtained and slope of the normal. Checks which of the given points satisfy the equation formed.

• Sine and cosine are complementary angles, i.e. $\sin \theta = \cos \left( {\dfrac{\pi }{2} - \theta } \right)$
• Tangent and cotangent are complementary angles, i.e. $\cot \theta = \tan \left( {\dfrac{\pi }{2} - \theta } \right)$
• Slope of tangent to the function $y = f(x)$ is given by $\dfrac{{dy}}{{dx}} = m$, where m is constant value
• Slope of normal to the function $y = f(x)$ is given by $\dfrac{{ - 1}}{{f'(x)}}$where $f'(x) = \dfrac{d}{{dx}}f(x)$

Complete step by step solution:
We are given $f(x) = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 + \sin x}}{{1 - \sin x}}} } \right),x \in \left( {0,\dfrac{\pi }{2}} \right)$..................… (1)
We find the value of the function at point $x = \dfrac{\pi }{6}$ by substituting the value of $x = \dfrac{\pi }{6}$in equation (1)
$\Rightarrow f(\dfrac{\pi }{6}) = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 + \sin \dfrac{\pi }{6}}}{{1 - \sin \dfrac{\pi }{6}}}} } \right)$
We know the value of $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$
$\Rightarrow f(\dfrac{\pi }{6}) = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 + \dfrac{1}{2}}}{{1 - \dfrac{1}{2}}}} } \right)$
Take LCM in both numerator and denominator of the fraction in RHS of the equation
$\Rightarrow f(\dfrac{\pi }{6}) = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{\dfrac{{2 + 1}}{2}}}{{\dfrac{{2 - 1}}{2}}}} } \right)$
$\Rightarrow f(\dfrac{\pi }{6}) = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{\dfrac{3}{2}}}{{\dfrac{1}{2}}}} } \right)$
Write fraction in RHS in simpler form
$\Rightarrow f(\dfrac{\pi }{6}) = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{3}{2} \times \dfrac{2}{1}} } \right)$
Cancel same factors from numerator and denominator in fraction in RHS of the equation
$\Rightarrow f(\dfrac{\pi }{6}) = {\tan ^{ - 1}}\left( {\sqrt 3 } \right)$
Since we know $\tan \dfrac{\pi }{3} = \sqrt 3$. Substitute the value of $\tan \dfrac{\pi }{3} = \sqrt 3$ in RHS
$\Rightarrow f(\dfrac{\pi }{6}) = {\tan ^{ - 1}}\left( {\tan \dfrac{\pi }{3}} \right)$
We cancel the function by its inverse in RHS of the equation
$\Rightarrow f(\dfrac{\pi }{6}) = \dfrac{\pi }{3}$
Since $y = f(x)$, therefore when $x = \dfrac{\pi }{6};y = \dfrac{\pi }{3}$
Point becomes $\left( {\dfrac{\pi }{6},\dfrac{\pi }{3}} \right)$...................… (2)
Now we know $f(x) = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 + \sin x}}{{1 - \sin x}}} } \right)$
Since we know sine and cosine are complementary angles, i.e. $\sin \theta = \cos \left( {\dfrac{\pi }{2} - \theta } \right)$
Substitute the value of $\sin x = \cos \left( {\dfrac{\pi }{2} - x} \right)$ in function
$\Rightarrow f(x) = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 + \cos \left( {\dfrac{\pi }{2} - x} \right)}}{{1 - \cos \left( {\dfrac{\pi }{2} - x} \right)}}} } \right)$..................… (3)
Now we know the identities $1 + \cos 2\theta = 2{\cos ^2}\theta$and $1 - \cos 2\theta = 2{\sin ^2}\theta$
If we put $\theta = \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)$
$\Rightarrow 1 + \cos 2\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right) = 2{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)$and$1 - \cos 2\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right) = 2{\sin ^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)$
Cancel 2 from the terms inside the angle
$\Rightarrow 1 + \cos \left( {\dfrac{\pi }{2} - x} \right) = 2{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)$and$1 - \cos \left( {\dfrac{\pi }{2} - x} \right) = 2{\sin ^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)$..................… (4)
Substitute values from equation (4) in equation (3)
$\Rightarrow f(x) = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{2{{\cos }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{2{{\sin }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}} } \right)$
Cancel same factor i.e. 2 from both numerator and denominator in the fraction
$\Rightarrow f(x) = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{{{\cos }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{{{\sin }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}} } \right)$ …………...… (5)
Since we know $\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta$, equation (5) becomes
$\Rightarrow f(x) = {\tan ^{ - 1}}\left( {\sqrt {{{\cot }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)} } \right)$
\Rightarrow f(x) = {\tan ^{ - 1}}\left( {\sqrt {{{\left\\{ {\cot \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)} \right\\}}^2}} } \right)
Cancel square root by square power in RHS of the equation
$\Rightarrow f(x) = {\tan ^{ - 1}}\left( {\cot \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)} \right)$......................… (6)
Since we know tangent and cotangent are complementary angles, i.e. $\cot \theta = \tan \left( {\dfrac{\pi }{2} - \theta } \right)$
Then the value of $\cot \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right) = \tan \left( {\dfrac{\pi }{2} - \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)} \right)$
i.e. $\cot \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right) = \tan \left( {\dfrac{\pi }{2} - \dfrac{\pi }{4} + \dfrac{x}{2}} \right)$
Take LCM of terms inside the bracket which are not associated with the variable ‘x’.
i.e.$\cot \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right) = \tan \left( {\dfrac{{2\pi - \pi }}{4} + \dfrac{x}{2}} \right)$
i.e. $\cot \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right) = \tan \left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right)$
Substitute this value of$\cot \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right) = \tan \left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right)$ in equation
$\Rightarrow f(x) = {\tan ^{ - 1}}\left[ {\tan \left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right)} \right]$
Cancel inverse of the function with the same function
$\Rightarrow f(x) = \left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right)$
Now we find slope of tangent of the function by differentiating it
$\Rightarrow \dfrac{d}{{dx}}f(x) = \dfrac{d}{{dx}}\left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right)$
$\Rightarrow f'(x) = \dfrac{d}{{dx}}\left( {\dfrac{\pi }{4}} \right) + \dfrac{d}{{dx}}\left( {\dfrac{x}{2}} \right)$
Use differentiation formula $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$
$\Rightarrow f'(x) = \dfrac{1}{2}$
So, slope of tangent i.e. $f'(x) = \dfrac{1}{2}$
We know slope of normal $= - 1/$ slope of tangent of function
So, slope of normal $= \dfrac{{ - 1}}{{\dfrac{1}{2}}} = - 2$
Now using the point $\left( {\dfrac{\pi }{6},\dfrac{\pi }{3}} \right)$and slope of tangent $- 2$ we form equation of tangent with the formula $\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)$
So, equation of normal to the function is $\left( {y - \dfrac{\pi }{3}} \right) = - 2\left( {x - \dfrac{\pi }{6}} \right)$
Take LCM on both side of the equation
$\Rightarrow \left( {\dfrac{{3y - \pi }}{3}} \right) = - 2\left( {\dfrac{{6x - \pi }}{6}} \right)$
Cancel same factors from numerator and denominator in RHS of the equation
$\Rightarrow \left( {\dfrac{{3y - \pi }}{3}} \right) = - \left( {\dfrac{{6x - \pi }}{3}} \right)$
Cancel same factors from denominators in both sides of the equation
$\Rightarrow 3y - \pi = - 6x + \pi$
Shift all values to left side of the equation
$\Rightarrow 6x + 3y - \pi - \pi = 0$
$\Rightarrow 6x + 3y - 2\pi = 0$.................… (7)
Now we check which of the given points (0,0)$$$$(0,\dfrac{{2\pi }}{3})$$$$(\dfrac{\pi }{6},0)$$$$(\dfrac{\pi }{4},0) satisfy equation (7)
Put $x = 0,y = 0$ in equation (7)
$\Rightarrow 6 \times 0 + 3 \times 0 - 2\pi = 0$
$\Rightarrow 0 - 2\pi = 0$
$\Rightarrow 2\pi = 0$
This is contradiction as $2\pi \ne 0$
Put $x = 0,y = \dfrac{{2\pi }}{3}$ in equation (7)
$\Rightarrow 6 \times 0 + 3 \times \dfrac{{2\pi }}{3} - 2\pi = 0$
$\Rightarrow 0 + 2\pi - 2\pi = 0$
$\Rightarrow 0 = 0$
This point satisfies the equation (7)
Put $x = \dfrac{\pi }{6},y = 0$ in equation (7)
$\Rightarrow 6 \times \dfrac{\pi }{6} + 3 \times 0 - 2\pi = 0$
$\Rightarrow \pi - 2\pi = 0$
$\Rightarrow - \pi = 0$
This is contradiction as $- \pi \ne 0$
Put $x = \dfrac{\pi }{4},y = 0$ in equation (7)
$\Rightarrow 6 \times \dfrac{\pi }{4} + 3 \times 0 - 2\pi = 0$
$\Rightarrow \dfrac{{3\pi }}{2} - 2\pi = 0$
Take LCM in LHS
$\Rightarrow \dfrac{{3\pi - 4\pi }}{2} = 0$
Cross multiply values from LHS to RHS
$\Rightarrow - \pi = 0$
This is contradiction as $- \pi \ne 0$
The normal to $y = f(x)$ at $x = \dfrac{\pi }{6}$ also passes through the point $\left( {0,\dfrac{{2\pi }}{3}} \right)$

$\therefore$Option B is correct.

Note: Many students make the mistake of rationalizing the term inside the inverse function which is not needed as we have to find the value of the inverse function at a given point which can be done by direct substitution. Also, many students use the formula of $\cot x = \dfrac{1}{{\tan x}}$ while removing the inverse function which is wrong as then we will have the value of function in reciprocal form and we will not be able to cancel inverse by the given function.