Question

If $5\cos^{- 1}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right) + 7\sin^{- 1}\left( {\dfrac{{2x}}{{\left( {1 + {x^2}} \right)}}} \right) – 4\tan^{- 1}\left( {\dfrac{{2x}}{{\left( {1 - {x^2}} \right)}}} \right) - \tan^{- 1}x = 5\pi $ , then what is the value of $x$?
A. $ - \sqrt 3 $
B. $\sqrt 2 $
C. 2
D. $\sqrt 3 $

Answer 1

Hint: First, substitute $x = \tan y$ in the given inverse trigonometric equation. Then simplify the equation using the trigonometric identities of $\sin2A, \cos2A$, and $\tan2A$. After that, use the inverse trigonometric identities and further simplify the equation. In the end, substitute the value of $y$ in the equation and simplify it to reach the required answer.

Formula Used:
$\sin2A = \dfrac{{2\tan A}}{{1 + \tan^{2}A}}$
$\cos2A = \dfrac{{1 - \tan^{2}A}}{{1 + \tan^{2}A}}$
$\tan2A = \dfrac{{2\tan A}}{{1 - \tan^{2}A}}$

Complete step by step solution:
The given trigonometric equation is $5\cos^{- 1}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right) + 7\sin^{- 1}\left( {\dfrac{{2x}}{{\left( {1 + {x^2}} \right)}}} \right) – 4\tan^{- 1}\left( {\dfrac{{2x}}{{\left( {1 - {x^2}} \right)}}} \right) - \tan^{- 1}x = 5\pi $.
Let’s simplify the above equation.
Let consider, $x = \tan y$
substitute $x = \tan y$ in the given equation.
$5\cos^{- 1}\left( {\dfrac{{1 - \tan^{2}y}}{{1 + \tan^{2}y}}} \right) + 7\sin^{- 1}\left( {\dfrac{{2\tan y}}{{\left( {1 + \tan^{2}y} \right)}}} \right) – 4\tan^{- 1}\left( {\dfrac{{2\tan y}}{{\left( {1 - \tan^{2}y} \right)}}} \right) - \tan^{- 1}\left( {\tan y} \right) = 5\pi $

Now apply the trigonometric identities of $\sin2A, \cos2A$, and $\tan2A$.
$5\cos^{- 1}\left( {\cos2y} \right) + 7\sin^{- 1}\left( {\sin2y} \right) – 4\tan^{- 1}\left( {\tan2y} \right) - \tan^{- 1}\left( {\tan y} \right) = 5\pi $
Apply the inverse trigonometric properties $\cos^{- 1}\left( {\cos A} \right) = A$, $\sin^{- 1}\left( {\sin A} \right) = A$, and$\tan^{- 1}\left( {\tan A} \right) = A$ in the above equation.
$5\left( {2y} \right) + 7\left( {2y} \right) - 4\left( {2y} \right) - \left( y \right) = 5\pi $
$ \Rightarrow 10y + 14y - 8y - y = 5\pi $
$ \Rightarrow 15y = 5\pi $
Divide both sides by $15$.
$y = \dfrac{\pi }{3}$

Now re-substitute the value of $y$ in the above equation.
$\tan^{- 1}x = \dfrac{\pi }{3}$
$ \Rightarrow x = \tan\dfrac{\pi }{3}$
$ \Rightarrow x = \sqrt 3 $ $\left[ {\because \tan\dfrac{\pi }{3} = \sqrt 3 } \right]$

Option ‘D’ is correct

Note: Students often try to differentiate the expression right from the beginning, which is technically correct but becomes excessively long and more prone to mistakes. To solve this type of question, first convert the equation into $\tan$ and simplify it using the trigonometric identities.