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Question:

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

Solution:

Disclaimer: The solution is provided for the following question.

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

Solution:

We know

$\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$, for all $x \in \mathrm{R}$

$\therefore \tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\cot ^{-1}\left(-\frac{1}{\sqrt{3}}\right)=\frac{\pi}{2}$        ...(1)

It is given that,

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

From (1) and (2), we get

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

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

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