Question

If $2\tan^{ - 1}\left( {\cos x} \right) = \tan^{ - 1}\left( {\csc^{2}x} \right)$, then what is the value of $x$?
A. $\dfrac{\pi }{2}$
B. $\pi $
C. $\dfrac{\pi }{6}$
D. $\dfrac{\pi }{3}$

Answer 1

Hint: First, simplify the left-hand side of the given equation using $2\tan^{ - 1}A$. Then, equate both sides and find a trigonometric equation. After that, solve the trigonometric equation by using the trigonometric identities and trigonometric ratio. In the end, solve the trigonometric equation to reach the required answer.

Formula Used:
$2\tan ^{\:-\:1}A\:=\:\tan ^{\:-\:1}\left(\:\frac{2A}{1\:-\:A^2}\:\right)$
$\cos^{2}A + \sin^{2}A = 1$
$\csc A = \dfrac{1}{{\sin A}}$

Complete step by step solution:
The given inverse trigonometric equation is $2\tan^{ - 1}\left( {\cos x} \right) = \tan^{ - 1}\left( {\csc^{2}x} \right)$.
Let’s simplify the left-hand side of the above equation by using the trigonometric identity $2\tan^{ - 1}A = \tan^{ - 1}\left( {\dfrac{{2A}}{{1 - {A^2}}}} \right)$.
$\tan^{ - 1}\left( {\dfrac{{2\cos x}}{{1 - \cos^{2}x}}} \right) = \tan^{ - 1}\left( {\csc^{2}x} \right)$
Now equate both sides.
We get,
$\frac{2\cos \:x}{1\:-\:\cos ^2x}\:=\:\csc ^2x$
Simplify the above equation using the formulas $\cos^{2}A + \sin^{2}A = 1$, and $\csc A = \dfrac{1}{{\sin A}}$.
$\dfrac{{2\cos x}}{{\cos^{2}x + \sin^{2}x - \cos^{2}x}} = \dfrac{1}{{\sin^{2}x}}$
$ \Rightarrow \dfrac{{2\cos x}}{{\sin^{2}x}} = \dfrac{1}{{\sin^{2}x}}$
$ \Rightarrow 2\cos x = 1$
Divide both sides by 2.
$\cos x = \dfrac{1}{2}$
$ \Rightarrow \cos x = \cos\dfrac{\pi }{3}$ $\left[ {\because \cos\dfrac{\pi }{3} = \dfrac{1}{2}} \right]$
$ \Rightarrow x = \dfrac{\pi }{3}$

Option ‘D’ is correct

Note: The identity $\sin^2x + \cos^2x = 1$ is useful for deriving many other important trigonometric identities too. It all depends on the right hand side of the identity to prove that with which trigonometric term, $\cos^2x$ or $\sin^2x$, the identity is to be used. Always remember trigonometry ratios and identities formulas.
Following are the basic trigonometric ratios:
$\csc A = \dfrac{1}{{\sin A}}$
$\sec A = \dfrac{1}{{\cos A}}$
$\cot A = \dfrac{1}{{\tan A}}$