The value of ^-1(√1+x^2+√1-x^2√1+x^2-√1-x^2), |x|

Question

The value of $	an^{-1}\left(\dfrac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}ight)$, $|x|<\dfrac12$, $x
e0$, is equal to

Accepted Answer

Let $x^2=\cos2\theta$, so $\theta=\dfrac12\cos^{-1}x^2$. Then $\sqrt{1+x^2}\pm\sqrt{1-x^2}$ relate to $\sqrt{1+\cos2\theta}\pm\sqrt{1-\cos2\theta}=\sqrt2(\cos\theta\pm\sin\theta)$:

$$\tan^{-1}\left(\frac{\sqrt{1+\cos2\theta}+\sqrt{1-\cos2\theta}}{\sqrt{1+\cos2\theta}-\sqrt{1-\cos2\theta}}\right)=\tan^{-1}\left(\frac{1+\tan\theta}{1-\tan\theta}\right)=\tan^{-1}\left(\tan\left(\frac{\pi}{4}+\theta\right)\right)=\frac{\pi}{4}+\theta.$$

$$=\frac{\pi}{4}+\frac12\cos^{-1}x^2.$$

Solutions · 1