Question

Tan inverse √1 + x square minus 1 by X in terms of tan inverse x

Answer 1

$\frac{1}{2}\tan^{-1}(x)$

Answer 2

Let $x = \tan \theta$. The expression transforms to $\tan^{-1}\left(\frac{\sec\theta - 1}{\tan\theta}\right)$. Using trigonometric identities, this simplifies to $\tan^{-1}\left(\frac{1-\cos\theta}{\sin\theta}\right)$. Applying half-angle formulas ($1-\cos\theta = 2\sin^2(\theta/2)$ and $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$), the expression becomes $\tan^{-1}(\tan(\theta/2))$. Since $\theta = \tan^{-1}(x)$, $\theta \in (-\pi/2, \pi/2)$, which implies $\theta/2 \in (-\pi/4, \pi/4)$. In this interval, $\tan^{-1}(\tan(\theta/2)) = \theta/2$. Substituting back $\theta = \tan^{-1}(x)$ yields the final result.