Question

If $\cos (2{\tan ^{ - 1}}x) = \dfrac{1}{2}$ then value of $x$ is:
A. $\sqrt 3 - 1$
B. $ \pm \sqrt 3 $
C. $ \pm \dfrac{1}{{\sqrt 3 }}$
D. $1 - \dfrac{1}{{\sqrt 3 }}$

Answer 1

Hint:
We know that $\cos 60^\circ = \dfrac{1}{2}$ so $2{\tan ^{ - 1}}x = 60^\circ $ then we get that ${\tan ^{ - 1}}x = 30^\circ $ now we know ${\tan ^{ - 1}}x$ is equal to any angle. So if
$
\Rightarrow {\tan ^{ - 1}}x = \theta \\
\Rightarrow x = \tan \theta \\
 $
Therefore we can solve easily for $x$

Complete step by step solution:
Here we are given the question $\cos (2{\tan ^{ - 1}}x) = \dfrac{1}{2}$ and here we are given the inverse of $\tan \theta $. So basically sine, cosine, tangent, cotangent, secant and cosecant are used to obtain an angle form any of the trigonometric ratios.
Here we used to write commonly that $\sin \theta = x$ then we can write that ${\sin ^{ - 1}}x = \theta $
So basically here $\theta {\text{ and }}x$ are the values of the angle and the numerical value.
So here if we are given either of them then we can use the inverse function in order to find another variable for that question and the most vital point in this is that the range of the inverse functions are the proper subsets of the domains of the original functions.
So here as we are given that
$\Rightarrow \cos (2{\tan ^{ - 1}}x) = \dfrac{1}{2}$
Now we can write it in the form of
$\Rightarrow 2{\tan ^{ - 1}}x = {\cos ^{ - 1}}\dfrac{1}{2}$
Now we know that the range of the ${\cos ^{ - 1}}x$ is between $0{\text{ and }}\dfrac{\pi }{2}$ and we know that $\cos 60^\circ = \dfrac{1}{2}$
So as we know that $\cos \dfrac{\pi }{3} = \dfrac{1}{2}$ we can write that
$\Rightarrow {\cos ^{ - 1}}\dfrac{1}{2} = \dfrac{\pi }{3}$
We can write that
$\Rightarrow 2{\tan ^{ - 1}}x = \dfrac{\pi }{3}$
$\Rightarrow {\tan ^{ - 1}}x = \dfrac{\pi }{6}$
Now we know that range for ${\tan ^{ - 1}}x$ is from $ - \dfrac{\pi }{2}{\text{ to }}\dfrac{\pi }{2}$
So we can write that
$\Rightarrow x = \pm \tan \dfrac{\pi }{6} \Rightarrow \pm \dfrac{1}{{\sqrt 3 }}$

So option C is our correct result.

Note:
We must know that for ${\sin ^{ - 1}}x,x$ must lie between $[ - 1,1]$ and so ${\sin ^{ - 1}}x$ is from $ - \dfrac{\pi }{2}{\text{ to }}\dfrac{\pi }{2}$
Similarly the domain of the ${\cos ^{ - 1}}x,x$ must lie between $[ - 1,1]$ and so ${\cos ^{ - 1}}x$ is from $ - \dfrac{\pi }{2}{\text{ to }}\dfrac{\pi }{2}$ which is its range and similarly we must know this for all the trigonometric functions.