Question

Prove that \[{{\tan }^{-1}}\left[ \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right]=\dfrac{\pi }{4}-\dfrac{1}{2}{{\cos }^{-1}}x,\dfrac{-1}{\sqrt{2}}\le x\le 1.\]

Answer 1

Hint: To simplify \[{{\tan }^{-1}}\left[ \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right]\] we put \[x=\cos 2\theta \] and then use \[1+\cos 2\theta =2{{\cos }^{2}}\theta \] and \[1-\cos 2\theta =2{{\sin }^{2}}\theta .\] We will simplify the given expression at last and we will compare all the terms using \[\dfrac{\tan A-\tan B}{1+\tan A\tan B}=\tan \left( A-B \right).\] At last, we will shift again \[x=\cos 2\theta \] to get the required solution.

Complete step by step answer:
We have \[{{\tan }^{-1}}\left[ \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right]\] and we have to show that the value is equal to \[\dfrac{\pi }{4}+\dfrac{1}{2}{{\cos }^{-1}}x.\] We will first simplify \[{{\tan }^{-1}}\left[ \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right]\] and then using the simplified term, we will solve further. We will start with the assumption that we have \[x=\cos 2\theta ,\] we will use this because \[1+\cos 2\theta =2{{\cos }^{2}}\theta \] and \[1-\cos 2\theta =2{{\sin }^{2}}\theta .\]
Now, we will put \[x=\cos 2\theta \] in \[\left[ \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right].\] We will get,
\[\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\dfrac{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}\]
As, \[1+\cos 2\theta =2{{\cos }^{2}}\theta \] and \[1-\cos 2\theta =2{{\sin }^{2}}\theta ,\] so we get,
\[\Rightarrow \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\dfrac{\sqrt{2{{\cos }^{2}}\theta }-\sqrt{2{{\sin }^{2}}\theta }}{\sqrt{2{{\cos }^{2}}\theta }+\sqrt{2{{\sin }^{2}}\theta }}\]
Simplifying, we get,
\[\Rightarrow \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\dfrac{\sqrt{2}\cos \theta -\sqrt{2}\sin \theta }{\sqrt{2}\cos \theta +\sqrt{2}\sin \theta }\]
Taking \[\sqrt{2}\] common, we get,
\[\Rightarrow \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\dfrac{\sqrt{2}}{\sqrt{2}}\left[ \dfrac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta } \right]\]
Diving the numerator and denominator by \[\cos \theta ,\] we get,
\[\Rightarrow \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\dfrac{\sqrt{2}}{\sqrt{2}}\left[ \dfrac{\dfrac{\cos \theta }{\cos \theta }-\dfrac{\sin \theta }{\cos \theta }}{\dfrac{\cos \theta }{\cos \theta }+\dfrac{\sin \theta }{\cos \theta }} \right]\]
Now simplifying using \[\dfrac{\sin \theta }{\cos \theta }=\tan \theta ,\] we will get,
\[\Rightarrow \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\dfrac{\sqrt{2}}{\sqrt{2}}\left[ \dfrac{1-\tan \theta }{1+\tan \theta } \right]\]
As \[\tan \dfrac{\pi }{4}=1,\] we will get,
\[\Rightarrow \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\left[ \dfrac{\tan \dfrac{\pi }{4}-\tan \theta }{1+\tan \dfrac{\pi }{4}.\tan \theta } \right]\]
Now, using \[\dfrac{\tan A-\tan B}{1+\tan A\tan B}=\tan \left( A-B \right),\] we will get,
\[\Rightarrow \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\tan \left( \dfrac{\pi }{4}-\theta \right)\]
So, we will get for \[x=\cos 2\theta \] as
\[\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\tan \left( \dfrac{\pi }{4}-\theta \right)\]
Now, applying \[{{\tan }^{-1}}\] on both the sides, we get,
\[\Rightarrow {{\tan }^{-1}}\left[ \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right]={{\tan }^{-1}}\left[ \tan \left( \dfrac{\pi }{4}-\theta \right) \right]\]
We know that, \[{{\tan }^{-1}}\left( \tan \theta \right)=\theta .\] So, we will get,
\[\Rightarrow {{\tan }^{-1}}\left[ \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right]=\dfrac{\pi }{4}-\theta \]
As, \[\cos 2\theta =x\]
\[\Rightarrow 2\theta ={{\cos }^{-1}}x\]
\[\Rightarrow \theta =\dfrac{1}{2}{{\cos }^{-1}}x\]
Using this in the above value, we will get,
\[\Rightarrow {{\tan }^{-1}}\left[ \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right]=\dfrac{\pi }{4}-\dfrac{1}{2}{{\cos }^{-1}}x\]

Note: We need to know certain formulas like \[1+\cos 2\theta =2{{\cos }^{2}}\theta ,\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta .\] Since, \[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta ,\] so \[\cos 2\theta ={{\cos }^{2}}\theta -\left( 1-{{\cos }^{2}}\theta \right).\]
Simplifying further, we get,
\[\Rightarrow \cos 2\theta ={{\cos }^{2}}\theta +{{\cos }^{2}}\theta -1\]
\[\Rightarrow 1+\cos 2\theta =2{{\cos }^{2}}\theta \]