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