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Question:

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

Solution:

We know that $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$.

We have

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

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

$\Rightarrow \tan ^{-1}(\sqrt{3})=\tan ^{-1} x$

$\Rightarrow x=\sqrt{3}$

$\therefore x=\sqrt{3}$

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