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Question: Prove that \[{{\tan }^{-1}}\left[ \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right]=\dfra...

Prove that tan⁡−1[1+x−1−x1+x+1−x]=π4−12cos⁡−1x,−12≤x≤1.{{\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.

Explanation

Solution

To simplify tan⁡−1[1+x−1−x1+x+1−x]{{\tan }^{-1}}\left[ \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right] we put x=cos⁡2θx=\cos 2\theta and then use 1+cos⁡2θ=2cos⁡2θ1+\cos 2\theta =2{{\cos }^{2}}\theta and 1−cos⁡2θ=2sin⁡2θ.1-\cos 2\theta =2{{\sin }^{2}}\theta . We will simplify the given expression at last and we will compare all the terms using tan⁡A−tan⁡B1+tan⁡Atan⁡B=tan⁡(A−B).\dfrac{\tan A-\tan B}{1+\tan A\tan B}=\tan \left( A-B \right). At last, we will shift again x=cos⁡2θx=\cos 2\theta to get the required solution.

Complete step by step answer:
We have tan⁡−1[1+x−1−x1+x+1−x]{{\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 π4+12cos⁡−1x.\dfrac{\pi }{4}+\dfrac{1}{2}{{\cos }^{-1}}x. We will first simplify tan⁡−1[1+x−1−x1+x+1−x]{{\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⁡2θ,x=\cos 2\theta , we will use this because 1+cos⁡2θ=2cos⁡2θ1+\cos 2\theta =2{{\cos }^{2}}\theta and 1−cos⁡2θ=2sin⁡2θ.1-\cos 2\theta =2{{\sin }^{2}}\theta .
Now, we will put x=cos⁡2θx=\cos 2\theta in [1+x−1−x1+x+1−x].\left[ \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right]. We will get,
1+x−1−x1+x+1−x=1+cos⁡2θ−1−cos⁡2θ1+cos⁡2θ+1−cos⁡2θ\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⁡2θ=2cos⁡2θ1+\cos 2\theta =2{{\cos }^{2}}\theta and 1−cos⁡2θ=2sin⁡2θ,1-\cos 2\theta =2{{\sin }^{2}}\theta , so we get,
⇒1+x−1−x1+x+1−x=2cos⁡2θ−2sin⁡2θ2cos⁡2θ+2sin⁡2θ\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,
⇒1+x−1−x1+x+1−x=2cos⁡θ−2sin⁡θ2cos⁡θ+2sin⁡θ\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 2\sqrt{2} common, we get,
⇒1+x−1−x1+x+1−x=22[cos⁡θ−sin⁡θcos⁡θ+sin⁡θ]\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⁡θ,\cos \theta , we get,
⇒1+x−1−x1+x+1−x=22[cos⁡θcos⁡θ−sin⁡θcos⁡θcos⁡θcos⁡θ+sin⁡θcos⁡θ]\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 sin⁡θcos⁡θ=tan⁡θ,\dfrac{\sin \theta }{\cos \theta }=\tan \theta , we will get,
⇒1+x−1−x1+x+1−x=22[1−tan⁡θ1+tan⁡θ]\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⁡π4=1,\tan \dfrac{\pi }{4}=1, we will get,
⇒1+x−1−x1+x+1−x=[tan⁡π4−tan⁡θ1+tan⁡π4.tan⁡θ]\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 tan⁡A−tan⁡B1+tan⁡Atan⁡B=tan⁡(A−B),\dfrac{\tan A-\tan B}{1+\tan A\tan B}=\tan \left( A-B \right), we will get,
⇒1+x−1−x1+x+1−x=tan⁡(π4−θ)\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⁡2θx=\cos 2\theta as
1+x−1−x1+x+1−x=tan⁡(π4−θ)\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\tan \left( \dfrac{\pi }{4}-\theta \right)
Now, applying tan⁡−1{{\tan }^{-1}} on both the sides, we get,
⇒tan⁡−1[1+x−1−x1+x+1−x]=tan⁡−1[tan⁡(π4−θ)]\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(tan⁡θ)=θ.{{\tan }^{-1}}\left( \tan \theta \right)=\theta . So, we will get,
⇒tan⁡−1[1+x−1−x1+x+1−x]=π4−θ\Rightarrow {{\tan }^{-1}}\left[ \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right]=\dfrac{\pi }{4}-\theta
As, cos⁡2θ=x\cos 2\theta =x
⇒2θ=cos⁡−1x\Rightarrow 2\theta ={{\cos }^{-1}}x
⇒θ=12cos⁡−1x\Rightarrow \theta =\dfrac{1}{2}{{\cos }^{-1}}x
Using this in the above value, we will get,
⇒tan⁡−1[1+x−1−x1+x+1−x]=π4−12cos⁡−1x\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⁡2θ=2cos⁡2θ,cos⁡2θ=cos⁡2θ−sin⁡2θ.1+\cos 2\theta =2{{\cos }^{2}}\theta ,\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta . Since, sin⁡2θ=1−cos⁡2θ,{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta , so cos⁡2θ=cos⁡2θ−(1−cos⁡2θ).\cos 2\theta ={{\cos }^{2}}\theta -\left( 1-{{\cos }^{2}}\theta \right).
Simplifying further, we get,
⇒cos⁡2θ=cos⁡2θ+cos⁡2θ−1\Rightarrow \cos 2\theta ={{\cos }^{2}}\theta +{{\cos }^{2}}\theta -1
⇒1+cos⁡2θ=2cos⁡2θ\Rightarrow 1+\cos 2\theta =2{{\cos }^{2}}\theta