Question

If $\alpha=2 \tan ^{-1}\left(\frac{1+x}{1-x}\right) \& \beta=\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ for $0<x<1$, then prove that $\alpha+\beta=\pi$. What the value of $\alpha+\beta$ will be if $x>1 ?$

Answer 1

SOLUTION : $\alpha=2 \tan ^{-1}\left(\frac{1-x}{1-x}\right) \quad ; \quad \beta=\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \\$

$\text { put } x=\tan \theta \quad\left\{\because x>1 \Rightarrow \theta-\frac{\pi}{4}\right\} \\$

$\Rightarrow \quad \alpha=2 \tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right) \quad \alpha=2 \tan ^{-1}\left\{\tan \left(\frac{\pi}{4}+\theta\right)\right\}=2\left\{\frac{\pi}{4}+\theta-\pi\right\}=2\left\{\theta-\frac{3 \pi}{4}\right\}=2 \theta-\frac{3 \pi}{2} \\$

$\Rightarrow \quad \beta=2 \sin ^{-1}\left\{\frac{1-\tan \tan ^2 \theta}{1+\tan ^2 \theta}\right\}=\sin ^{-1}(\cos 2 \theta)=\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-2 \theta\right)\right]=\frac{\pi}{2}-2 \theta$