^ -1 (1-x^2 2 x )+ ^ -1 (1-x^2 1+x^2 ) is equal to - Vidyadip

MCQ

$\sin \left\{\tan ^{-1}\left(\frac{1-x^2}{2 x}\right)+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right\}$ is equal to

A
$0$
✓
1
C
$\sqrt{2}$
D
$\frac{1}{\sqrt{2}}$

✓

Answer

Correct option: B.

1

(B) Putting $x=\tan \theta$, we get
$\sin \left[\tan ^{-1}\left(\frac{1-x^2}{2 x}\right)+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]$
$=\sin \left[\tan ^{-1}\left(\frac{1-\tan ^2 \theta}{2 \tan \theta}\right)+\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)\right]$
$\begin{array}{l}=\sin \left[\tan ^{-1}(\cot 2 \theta)+\cos ^{-1}(\cos 2 \theta)\right] \\ =\sin \left[\tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-2 \theta\right)\right\}+\cos ^{-1}(\cos 2 \theta)\right]\end{array}$
$=\sin \frac{\pi}{2}=1$

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