Solve the equation: $ 2{\tan ^{ - 1}}\left( {\cos x} \right) = {\tan ^{ - 1}}\left( {2\cos ecx} \right) $ ?

Hint: In the question, we have to solve the inverse trigonometric equation given to us. We can solve this by taking the term inside the trigonometric function as an angle. One must have a strong grip over the concepts of trigonometry and algebra. Then, we have the tangent double angle formula that is \[\tan \left( {2\theta } \right) = \left( {\dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}} \right)\]. Then substituting the value of theta we can get the desired result.

Complete step-by-step answer:
In the given question, we need to solve the equation $ 2{\tan ^{ - 1}}\left( {\cos x} \right) = {\tan ^{ - 1}}\left( {2\cos ecx} \right) $ .
Now, to solve the equation consisting of inverse trigonometric functions given to us, we will have to take tangent functions on both sides of the equation. So, we get,
 $ \Rightarrow \tan \left( {2{{\tan }^{ - 1}}\left( {\cos x} \right)} \right) = \tan \left( {{{\tan }^{ - 1}}\left( {2\cos ecx} \right)} \right) $
Now, we know that the value of \[\tan \left( {{{\tan }^{ - 1}}x} \right) = x\] for every real value of x. Also, we take $ {\tan ^{ - 1}}\left( {\cos x} \right) $ as t. So, we get,
 $ \Rightarrow \tan \left( {2{{\tan }^{ - 1}}\left( {\cos x} \right)} \right) = 2\cos ecx $
 $ \Rightarrow \tan \left( {2t} \right) = 2\cos ecx $
Now, using the trigonometric formula \[\tan \left( {2\theta } \right) = \left( {\dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}} \right)\] in the left side of the equation, we get,
 $ \Rightarrow \dfrac{{2\tan t}}{{1 - {{\tan }^2}t}} = 2\cos ecx $
Now, substituting the value of t in the equation, we get,
 $ \Rightarrow \dfrac{{2\tan \left( {{{\tan }^{ - 1}}\left( {\cos x} \right)} \right)}}{{1 - {{\tan }^2}\left( {{{\tan }^{ - 1}}\left( {\cos x} \right)} \right)}} = 2\cos ecx $
Now, again using the fact that the value of \[\tan \left( {{{\tan }^{ - 1}}x} \right)\] equals x for every real value of x.
So, we get,
 $ \Rightarrow \dfrac{{2\cos x}}{{1 - {{\left( {\cos x} \right)}^2}}} = 2\cos ecx $
Now, using the trigonometric formulae $ \cos ecx = \dfrac{1}{{\sin x}} $ and the trigonometric identity $ {\sin ^2}x + {\cos ^2}x = 1 $ , we get,
 $ \Rightarrow \dfrac{{2\cos x}}{{1 - {{\left( {\cos x} \right)}^2}}} = \dfrac{2}{{\sin x}} $
Simplifying the expression further, we get,
 $ \Rightarrow \dfrac{{\cos x}}{{{{\sin }^2}x}} = \dfrac{1}{{\sin x}} $
Cancelling the common factors in numerator and denominator, we get,
 $ \Rightarrow \dfrac{{\cos x}}{{\sin x}} = 1 $
Now, we know that $ \cot x = \dfrac{{\cos x}}{{\sin x}} $ . Hence, we get,
 $ \Rightarrow \cot x = 1 $
Now, we know that cotangent and tangent are reciprocal ratios of each other. Hence, we can write $ \cot x = \dfrac{1}{{\tan x}} $ . So, we get,
 $ \Rightarrow \tan x = 1 $
Now, we know that the general solution of a trigonometric equation $ \tan x = \tan y $ can be written as $ x = n\pi + y $ .
So, we know that the value of $ \tan \dfrac{\pi }{4} $ is one. So, we get the solution of the equation as,
 $ \Rightarrow x = n\pi + \dfrac{\pi }{4} $
Hence, we get the solution of the trigonometric equation $ 2{\tan ^{ - 1}}\left( {\cos x} \right) = {\tan ^{ - 1}}\left( {2\cos ecx} \right) $ as $ x = n\pi + \dfrac{\pi }{4} $ .

Note: In the above problem we have arctangent and we call it the inverse of the tangent. Tangent and inverse of tangent cancels out, we have: $ \tan \left( {\arctan x} \right) = \tan \left( {{{\tan }^{ - 1}}x} \right) = x $ . But this is not the case in the given problem. Because we have two multiplied to the arctangent on the left side of the equation, we have to employ the double angle formula for calculating the tangent of the angle.