Question

If ${\tan }^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)$ = α, then x² =
(a) sin 2 α
(b) sin α
(c) cos 2 α
(d) cos α

Answer 1

(a) sin 2α

$$\begin{gathered} {\tan }^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right)=\alpha \\ \Rightarrow \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}=\tan \alpha \\ \Rightarrow \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\times \frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=\tan \alpha \\ \Rightarrow \frac{{\left(\sqrt{1+x^2}\right)}^2+{\left(\sqrt{1-x^2}\right)}^2-2\sqrt{1+x^2}\sqrt{1-x^2}}{{\left(\sqrt{1+x^2}\right)}^2-{\left(\sqrt{1-x^2}\right)}^2}=\tan \alpha \\ \Rightarrow \frac{1-\sqrt{1-x^4}}{x^2}=\tan \alpha \\ \Rightarrow x^2\tan \alpha =1-\sqrt{1-x^4} \\ \Rightarrow \sqrt{1-x^4}=1-x^2\tan \alpha \\ \Rightarrow 1-x^4=1+x^4{\tan }^2\alpha -2x^2\tan \alpha \\ \Rightarrow x^4+x^4{\tan }^2\alpha -2x^2\tan \alpha =0 \\ \Rightarrow x^4{\sec }^2\alpha -2x^2\tan \alpha =0 \\ \Rightarrow x^2\left(x^2{\sec }^2\alpha -2\tan \alpha \right)=0 \\ \Rightarrow x^2{\sec }^2\alpha -2\tan \alpha =0\left[\because x^2\ne 0\right] \\ \Rightarrow x^2{\sec }^2\alpha =2\tan \alpha \\ \Rightarrow x^2=\frac{2\tan \alpha }{{\sec }^2\alpha }=2\sin \alpha \cos \alpha =\sin 2\alpha \end{gathered}$$