Prove that [ ^ -1 (1-x^2 2 x )+ ^ -1 (1-x^2 1+x^2 ) ]=1 - Vidyadip

Question

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

✓

Answer

$\text { L.H.S. }=\sin \left[\tan ^{-1}\left(\frac{1-x^2}{2 x}\right)+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]$
Substituting $x=\tan \theta$, we get
$ \text { L.H.S. }=\sin \left[\tan ^{-1}\left(\frac{1-\tan ^2 \theta}{2 \tan \theta}\right)+\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)\right]$
$=\sin \left[\tan ^{-1}\left(\frac{1}{\tan 2} \theta\right)+\cos ^{-1}(\cos 2 \theta)\right]$
$=\sin \left[\tan ^{-1}(\cot 2 \theta)+\cos ^{-1}(\cos 2 \theta)\right]$
$=\sin \left[\tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-2 \theta\right)\right\}+2 \theta\right]$
$=\sin \left(\frac{\pi}{2}-2 \theta+2 \theta\right)$
$=\sin \left(\frac{\pi}{2}\right)$
$=1$
$=\text { R.H.S. } $

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