Question

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

Answer 1

Hint: Given equation is \[2{{\tan }^{-1}}x={{\cos }^{-1}}\left(\dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\]. We have to show that L.H.S = R.H.S. We have to prove the above equation. Consider \[x=\tan \theta \] 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 \[{{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\]
We have to substitute the value of \[x=\tan \theta \] in the above term.
After substituting the value \[x=\tan \theta \] the term further appeared as follows: \[{{\cos }^{-1}}\left( \dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right)\]
We know that \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]. Substituting this value in the above term, the term further appeared as follows:
As we all know that \[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\], substituting this in the above term, the term further appeared as \[{{\cos }^{-1}}\left( \dfrac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{{{\sec }^{2}}\theta \times {{\cos }^{2}}\theta } \right)\]
We all know that \[{{\sec }^{2}}\theta \]=\[\dfrac{1}{{{\cos }^{2}}\theta }\]
The term now appears as \[{{\cos }^{-1}}\left( \dfrac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{\dfrac{1}{{{\cos }^{2}}\theta }\times {{\cos }^{2}}\theta } \right)\]
Cancelling both the terms in the denominator further leads to the term as \[{{\cos }^{-1}}\left( {{\cos }^{2}}\theta -{{\sin }^{2}}\theta \right)\]
We all know that \[\cos 2\theta \]=\[{{\cos }^{2}}\theta -{{\sin }^{2}}\theta \]
By substituting this value in the above term leads the equation to \[{{\cos }^{-1}}\left( \cos 2\theta \right)\]
We know that \[{{\cos }^{-1}}\left( \cos 2\theta \right)\]=\[2\theta \]
The term appeared is \[2\theta \]
At first we substituted the value \[x=\tan \theta \]. Now substituting the value of \[\theta \] in the above term, the equation now appears as, \[\theta ={{\tan }^{-1}}x\]
\[\Rightarrow \]2 \[{{\tan }^{-1}}x\]
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 \[2{{\tan }^{-1}}x={{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\]
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 \[x=\tan \theta \] is the main step in the above solution.