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

If cos⁡(2tan⁡−1x)=12\cos (2{\tan ^{ - 1}}x) = \dfrac{1}{2} then value of xx is:
A. 3−1\sqrt 3 - 1
B. ±3\pm \sqrt 3
C. ±13 \pm \dfrac{1}{{\sqrt 3 }}
D. 1−131 - \dfrac{1}{{\sqrt 3 }}

Explanation

Solution

We know that cos⁡60∘=12\cos 60^\circ = \dfrac{1}{2} so 2tan⁡−1x=60∘2{\tan ^{ - 1}}x = 60^\circ then we get that tan⁡−1x=30∘{\tan ^{ - 1}}x = 30^\circ now we know tan⁡−1x{\tan ^{ - 1}}x is equal to any angle. So if
⇒tan⁡−1x=θ ⇒x=tan⁡θ  \Rightarrow {\tan ^{ - 1}}x = \theta \\\ \Rightarrow x = \tan \theta \\\
Therefore we can solve easily for xx

Complete step by step solution:
Here we are given the question cos⁡(2tan⁡−1x)=12\cos (2{\tan ^{ - 1}}x) = \dfrac{1}{2} and here we are given the inverse of tan⁡θ\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⁡θ=x\sin \theta = x then we can write that sin⁡−1x=θ{\sin ^{ - 1}}x = \theta
So basically here θ and x\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
⇒cos⁡(2tan⁡−1x)=12\Rightarrow \cos (2{\tan ^{ - 1}}x) = \dfrac{1}{2}
Now we can write it in the form of
⇒2tan⁡−1x=cos⁡−112\Rightarrow 2{\tan ^{ - 1}}x = {\cos ^{ - 1}}\dfrac{1}{2}
Now we know that the range of the cos⁡−1x{\cos ^{ - 1}}x is between 0 and π20{\text{ and }}\dfrac{\pi }{2} and we know that cos⁡60∘=12\cos 60^\circ = \dfrac{1}{2}
So as we know that cos⁡π3=12\cos \dfrac{\pi }{3} = \dfrac{1}{2} we can write that
⇒cos⁡−112=π3\Rightarrow {\cos ^{ - 1}}\dfrac{1}{2} = \dfrac{\pi }{3}
We can write that
⇒2tan⁡−1x=π3\Rightarrow 2{\tan ^{ - 1}}x = \dfrac{\pi }{3}
⇒tan⁡−1x=π6\Rightarrow {\tan ^{ - 1}}x = \dfrac{\pi }{6}
Now we know that range for tan⁡−1x{\tan ^{ - 1}}x is from −π2 to π2 - \dfrac{\pi }{2}{\text{ to }}\dfrac{\pi }{2}
So we can write that
⇒x=±tan⁡π6⇒±13\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⁡−1x,x{\sin ^{ - 1}}x,x must lie between [−1,1][ - 1,1] and so sin⁡−1x{\sin ^{ - 1}}x is from −π2 to π2 - \dfrac{\pi }{2}{\text{ to }}\dfrac{\pi }{2}
Similarly the domain of the cos⁡−1x,x{\cos ^{ - 1}}x,x must lie between [−1,1][ - 1,1] and so cos⁡−1x{\cos ^{ - 1}}x is from −π2 to π2 - \dfrac{\pi }{2}{\text{ to }}\dfrac{\pi }{2} which is its range and similarly we must know this for all the trigonometric functions.