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AI-তৈরি নমুনা উত্তর

Given equation:
$2\tan^{-1} \left(\frac{x-1}{x+1}\right) = \tan^{-1} \frac{1}{\sqrt{3}}$

Since $\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$:
$2\tan^{-1} \left(\frac{x-1}{x+1}\right) = \frac{\pi}{6}$
$\tan^{-1} \left(\frac{x-1}{x+1}\right) = \frac{\pi}{12}$
$\frac{x-1}{x+1} = \tan 15^\circ$

We know that $\tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1} = 2 - \sqrt{3}$.

Therefore,
$\frac{x-1}{x+1} = 2 - \sqrt{3}$
$x - 1 = (2 - \sqrt{3})(x + 1)$
$x - 1 = (2 - \sqrt{3})x + 2 - \sqrt{3}$
$x - (2 - \sqrt{3})x = 3 - \sqrt{3}$
$x(\sqrt{3} - 1) = \sqrt{3}(\sqrt{3} - 1)$
$x = \sqrt{3}$