If we are given $ 2{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( {{\operatorname{cosec}}^{2}}x \right) $ , then find the value of x.
(a) $ \dfrac{\pi }{2} $ .
(b) $ \pi $ .
(c) $ \dfrac{\pi }{6} $ .
(d) $ \dfrac{\pi }{3} $ .

Hint: We convert $ 2{{\tan }^{-1}}\left( \cos x \right) $ to $ {{\tan }^{-1}}\left( \cos x \right)+{{\tan }^{-1}}\left( \cos x \right) $ for applying $ {{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right) $ . After applying the formula, we obtain a value for $ \cos x $ . Using the value of $ \cos x $ , we find the value of x. We verify by substituting the value of ‘x’ in the original given equation.

Complete step-by-step answer:
We are given $ 2{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( {{\operatorname{cosec}}^{2}}x \right) $ , and we need to find the value of ‘x’.
We have $ {{\tan }^{-1}}\left( \cos x \right)+{{\tan }^{-1}}\left( \cos x \right)={{\tan }^{-1}}\left( {{\operatorname{cosec}}^{2}}x \right) $ .
We know that $ {{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right) $ for ab<1.
We know the value of cos x lies in the interval $ \left[ -1,1 \right] $ and the value of $ \cos x\times \cos x $ lies in the interval $ \left[ 0,1 \right] $ .
So, we have $ {{\tan }^{-1}}\left( \dfrac{\cos x+\cos x}{1-\left( \cos x\times \cos x \right)} \right)={{\tan }^{-1}}\left( {{\operatorname{cosec}}^{2}}x \right) $ .
We have $ {{\tan }^{-1}}\left( \dfrac{2\cos x}{1-{{\cos }^{2}}x} \right)={{\tan }^{-1}}\left( {{\operatorname{cosec}}^{2}}x \right) $ .
We know from the trigonometric identity $ {{\sin }^{2}}x=1-{{\cos }^{2}}x $ .
We have $ {{\tan }^{-1}}\left( \dfrac{2\cos x}{{{\sin }^{2}}x} \right)={{\tan }^{-1}}\left( {{\operatorname{cosec}}^{2}}x \right) $ .
We know that $ \operatorname{cosec}x=\dfrac{1}{\sin x} $ .
We have $ {{\tan }^{-1}}\left( 2\cos x.{{\operatorname{cosec}}^{2}}x \right)={{\tan }^{-1}}\left( {{\operatorname{cosec}}^{2}}x \right) $ .
We know that if $ {{\tan }^{-1}}\left( x \right)={{\tan }^{-1}}\left( y \right) $ , then x = y.
We have $ 2\cos x.{{\operatorname{cosec}}^{2}}x={{\operatorname{cosec}}^{2}}x $ .
We have 2cos x = 1.
We have $ \cos x=\dfrac{1}{2} $ .
We have \[x={{\cos }^{-1}}\left( \dfrac{1}{2} \right)\].
We have $ x={{60}^{o}} $ .
We know that $ \pi ={{180}^{o}} $ .
We have $ x={{60}^{o}}\times \dfrac{{{180}^{o}}}{{{180}^{o}}} $ .
We have \[x=\dfrac{\pi }{3}\].
We got the value of ‘x’ as $ \dfrac{\pi }{3} $ .
Now, we verify the obtained result.
We have $ 2{{\tan }^{-1}}\left( \cos \left( \dfrac{\pi }{3} \right) \right)={{\tan }^{-1}}\left( {{\operatorname{cosec}}^{2}}\left( \dfrac{\pi }{3} \right) \right) $ .
We know that $ \cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2} $ and $ \operatorname{cosec}\left( \dfrac{\pi }{3} \right)=\dfrac{2}{\sqrt{3}} $ .
So, we have $ 2{{\tan }^{-1}}\left( \dfrac{1}{2} \right)={{\tan }^{-1}}\left( {{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}} \right) $ .
We have $ {{\tan }^{-1}}\left( \dfrac{1}{2} \right)+{{\tan }^{-1}}\left( \dfrac{1}{2} \right)={{\tan }^{-1}}\left( \dfrac{4}{3} \right) $ .
Since, $ \dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1}{4} $ , which is less than 1, we use $ {{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right) $ .
We have $ {{\tan }^{-1}}\left( \dfrac{\dfrac{1}{2}+\dfrac{1}{2}}{1-\left( \dfrac{1}{2}\times \dfrac{1}{2} \right)} \right)={{\tan }^{-1}}\left( \dfrac{4}{3} \right) $ .
We have $ {{\tan }^{-1}}\left( \dfrac{1}{1-\left( \dfrac{1}{4} \right)} \right)={{\tan }^{-1}}\left( \dfrac{4}{3} \right) $ .
We have $ {{\tan }^{-1}}\left( \dfrac{1}{\dfrac{3}{4}} \right)={{\tan }^{-1}}\left( \dfrac{4}{3} \right) $ .
We have $ {{\tan }^{-1}}\left( \dfrac{4}{3} \right)={{\tan }^{-1}}\left( \dfrac{4}{3} \right) $ .
∴ We have proved that the value of ‘x’ is $ \dfrac{\pi }{3} $ .
So, the correct answer is “Option D”.

Note: We should not every time use $ {{\tan }^{-1}}\left( a \right)+{{\tan }^{-1}}\left( b \right)={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right) $ as it depends on the value of ab. We always verify the value obtained if the given equation of inverse trigonometric functions consists of arctanx $ \left( {{\tan }^{-1}}x \right) $ . Sometimes, the obtained value may not satisfy the given equation in such cases we declare that ‘x’ doesn’t have a solution.