question_answer

If \[{{\tan }^{-1}}x-{{\cot }^{-1}}x={{\tan }^{-1}}\frac{1}{\sqrt{3}},\] then find the value of x.

Answer:

We have, \[{{\tan }^{-1}}x-{{\cot }^{-1}}x={{\tan }^{-1}}\frac{1}{\sqrt{3}}\] \[\Rightarrow \]   \[{{\tan }^{-1}}x-{{\cot }^{-1}}x=\frac{\pi }{6}\]                    ?(i) Also, we have \[{{\tan }^{-1}}x+{{\cot }^{-1}}x=\frac{\pi }{2}\]         ?(ii) On adding Eqs. (i) and (ii), we get \[2{{\tan }^{-1}}x=\frac{\pi }{6}+\frac{\pi }{2}\] \[\Rightarrow \] \[2{{\tan }^{-1}}x=\frac{2\pi }{3}\] \[\Rightarrow \] \[{{\tan }^{-1}}x=\frac{\pi }{3}\] \[\Rightarrow \] \[x=\tan \frac{\pi }{3}=\sqrt{3}\]