Question

The given equation is $\tan ^{-1}\left(\frac{1+x}{1-x}\right)=4\tan ^{-1}x$ solve x

Answer 1

2-\sqrt{3}

Answer 2

Let $\theta = \tan^{-1}x$. Then $x = \tan\theta$. The equation becomes $\tan^{-1}\left(\frac{1+\tan\theta}{1-\tan\theta}\right) = 4\theta$. Using the identity $\tan(\frac{\pi}{4}+\theta) = \frac{1+\tan\theta}{1-\tan\theta}$, we have $\tan^{-1}(\tan(\frac{\pi}{4}+\theta)) = 4\theta$. This implies $\frac{\pi}{4}+\theta - n\pi = 4\theta$ for some integer $n$. So, $3\theta = \frac{\pi}{4} - n\pi$, which gives $\theta = \frac{\pi}{12} - \frac{n\pi}{3}$. Since the range of $\tan^{-1}y$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$, we have $-\frac{\pi}{2} < 4\theta < \frac{\pi}{2}$, so $-\frac{\pi}{8} < \theta < \frac{\pi}{8}$. For $n=0$, $\theta = \frac{\pi}{12}$, which lies in $(-\frac{\pi}{8}, \frac{\pi}{8})$. For other integer values of $n$, $\theta$ is outside this range. Thus, $x = \tan(\frac{\pi}{12}) = \tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{(\sqrt{3}-1)^2}{3-1} = \frac{3 - 2\sqrt{3} + 1}{2} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$.