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Question: How do you simplify \( \cos \left( {2{{\tan }^{ - 1}}x} \right) \) ?...

How do you simplify cos⁡(2tan⁡−1x)\cos \left( {2{{\tan }^{ - 1}}x} \right) ?

Explanation

Solution

Hint : We can solve this by taking θ=tan⁡−1(x)\theta = {\tan ^{ - 1}}\left( x \right) . Then, we have the tangent double angle formula that is tan⁡(2θ)=(2tan⁡θ1−tan⁡2θ)\tan \left( {2\theta } \right) = \left( {\dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}} \right) and cosine double angle formula cos⁡(2θ)=(1−tan⁡2θ1+tan⁡2θ)\cos \left( {2\theta } \right) = \left( {\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right). Then substituting the value of theta we can get the desired result. We can do this directly by without taking θ=tan⁡−1(x)\theta = {\tan ^{ - 1}}\left( x \right) and substituting in the formula but it will be a little bit confusing while substituting in the formula.

Complete step by step solution:
In the given question, we need to simplify the expression cos⁡(2tan⁡−1x)\cos \left( {2{{\tan }^{ - 1}}x} \right) .
Let’s take θ=tan⁡−1(x)\theta = {\tan ^{ - 1}}\left( x \right). Then the given problem becomes,
cos⁡(2tan⁡−1x)=cos⁡(2θ)\cos \left( {2{{\tan }^{ - 1}}x} \right) = \cos \left( {2\theta } \right)
We know the cosine double angle formula that is cos⁡(2θ)=(1−tan⁡2θ1+tan⁡2θ)\cos \left( {2\theta } \right) = \left( {\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right).
But we have taken θ=tan⁡−1(x)\theta = {\tan ^{ - 1}}\left( x \right). Substituting this we have,
⇒cos⁡(2θ)=(1−tan⁡2θ1+tan⁡2θ)\Rightarrow \cos \left( {2\theta } \right) = \left( {\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)
⇒cos⁡(2θ)=(1−tan⁡2(tan⁡−1x)1+tan⁡2(tan⁡−1x))\Rightarrow \cos \left( {2\theta } \right) = \left( {\dfrac{{1 - {{\tan }^2}\left( {{{\tan }^{ - 1}}x} \right)}}{{1 + {{\tan }^2}\left( {{{\tan }^{ - 1}}x} \right)}}} \right)
We know that tan⁡(tan⁡−1x)=x\tan \left( {{{\tan }^{ - 1}}x} \right) = x .
Tangent and arctangent will get cancel in the numerator and denominator, so we have,
⇒cos⁡(2θ)=(1−x21+x2)\Rightarrow \cos \left( {2\theta } \right) = \left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)
Hence, we get cos⁡(2tan⁡−1x)\cos \left( {2{{\tan }^{ - 1}}x} \right) as (1−x21+x2)\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right).
So, the correct answer is “(1−x21+x2)\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)”.

Note : If they give us to evaluate tan⁡(tan⁡−1x)\tan \left( {{{\tan }^{ - 1}}x} \right) . Then it is simple to solve.
Since arctangent is inverse of tangent then we have,
tan⁡(arctan⁡x)=tan⁡(tan⁡−1x)\tan \left( {\arctan x} \right) = \tan \left( {{{\tan }^{ - 1}}x} \right)
Tangent and inverse of tangent cancels out, we have: tan⁡(arctan⁡x)=tan⁡(tan⁡−1x)=x\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 and it is inside the angle of tangent, we have to find the value of cosine of the double angle.
In the above problem we have arctangent and we call it as the inverse of the tangent. That is tan⁡(arctan⁡x)=tan⁡(tan⁡−1x)\tan \left( {\arctan x} \right) = \tan \left( {{{\tan }^{ - 1}}x} \right) . Cosine double angle formula has three forms and we can use all of them depending upon the situation and problem.