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Question: Prove that \[2{{\tan }^{-1}}x={{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\]...

Prove that 2tan⁡−1x=cos⁡−1(1−x21+x2)2{{\tan }^{-1}}x={{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)

Explanation

Solution

Hint: Given equation is 2tan⁡−1x=cos⁡−1(1−x21+x2)2{{\tan }^{-1}}x={{\cos }^{-1}}\left(\dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right). We have to show that L.H.S = R.H.S. We have to prove the above equation. Consider x=tan⁡θx=\tan \theta and solve the terms using the trigonometric formulas and certain mathematical operations to arrive at the solution.
Complete step-by-step answer:
Now considering the given term cos⁡−1(1−x21+x2){{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)
We have to substitute the value of x=tan⁡θx=\tan \theta in the above term.
After substituting the value x=tan⁡θx=\tan \theta the term further appeared as follows: cos⁡−1(1−tan⁡2θ1+tan⁡2θ){{\cos }^{-1}}\left( \dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right)
We know that tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta }. Substituting this value in the above term, the term further appeared as follows:
As we all know that sec⁡2θ−tan⁡2θ=1{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1, substituting this in the above term, the term further appeared as cos⁡−1(cos⁡2θ−sin⁡2θsec⁡2θ×cos⁡2θ){{\cos }^{-1}}\left( \dfrac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{{{\sec }^{2}}\theta \times {{\cos }^{2}}\theta } \right)
We all know that sec⁡2θ{{\sec }^{2}}\theta =1cos⁡2θ\dfrac{1}{{{\cos }^{2}}\theta }
The term now appears as cos⁡−1(cos⁡2θ−sin⁡2θ1cos⁡2θ×cos⁡2θ){{\cos }^{-1}}\left( \dfrac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{\dfrac{1}{{{\cos }^{2}}\theta }\times {{\cos }^{2}}\theta } \right)
Cancelling both the terms in the denominator further leads to the term as cos⁡−1(cos⁡2θ−sin⁡2θ){{\cos }^{-1}}\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)
We all know that cos⁡2θ\cos 2\theta =cos⁡2θ−sin⁡2θ{{\cos }^{2}}\theta -{{\sin }^{2}}\theta
By substituting this value in the above term leads the equation to cos⁡−1(cos⁡2θ){{\cos }^{-1}}\left( \cos 2\theta \right)
We know that cos⁡−1(cos⁡2θ){{\cos }^{-1}}\left( \cos 2\theta \right)=2θ2\theta
The term appeared is 2θ2\theta
At first we substituted the value x=tan⁡θx=\tan \theta . Now substituting the value of θ\theta in the above term, the equation now appears as, θ=tan⁡−1x\theta ={{\tan }^{-1}}x
⇒\Rightarrow 2 tan⁡−1x{{\tan }^{-1}}x
The obtained solution is the term which is on the L.H.S. Hence we have to show that L.H.S = R.H.S.
Hence proved that 2tan⁡−1x=cos⁡−1(1−x21+x2)2{{\tan }^{-1}}x={{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)
Note: For the above problem to solve we must have all the knowledge regarding trigonometric formulas, their substitutions, identities etc. If one formula is missed we cannot arrive at the solution.
Considering x=tan⁡θx=\tan \theta is the main step in the above solution.