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Question:

If $\tan ^{-1} x-\cot ^{-1} x=\tan ^{-1} \sqrt{3}$, then $x=$ _______________________.

Solution:

$\tan ^{-1} x-\cot ^{-1} x=\tan ^{-1} \sqrt{3}$

$\Rightarrow \tan ^{-1} x-\left(\frac{\pi}{2}-\tan ^{-1} x\right)=\frac{\pi}{3}$                   $\left(\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right)$

$\Rightarrow 2 \tan ^{-1} x=\frac{\pi}{2}+\frac{\pi}{3}$

$\Rightarrow \tan ^{-1} x=\frac{\pi}{4}+\frac{\pi}{6}$

$\Rightarrow x=\tan \left(\frac{\pi}{4}+\frac{\pi}{6}\right)$

$\Rightarrow x=\frac{\tan \frac{\pi}{4}+\tan \frac{\pi}{6}}{1-\tan \frac{\pi}{4} \times \tan \frac{\pi}{6}}$

$\Rightarrow x=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$

$\Rightarrow x=\frac{(\sqrt{3}+1)^{2}}{(\sqrt{3}-1)(\sqrt{3}+1)}$

$\Rightarrow x=\frac{4+2 \sqrt{3}}{2}$

$\Rightarrow x=2+\sqrt{3}$

If $\tan ^{-1} x-\cot ^{-1} x=\tan ^{-1} \sqrt{3}$, then $x=\frac{2+\sqrt{3}}{.}$

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