Solve for `x :` `cot^(-1)x-cot^(-1)(x+2)=pi/(12)` , where `x >0`

Answer

`x= sqrt{3}`

Explanation

\begin{aligned}
&\Rightarrow \cot ^{-1}(x)-\cot ^{-1}(x+2)=\frac{\pi}{12}\\
&\Rightarrow \tan ^{-1}\left(\frac{1}{x}\right)+\cot ^{-1}\left(\frac{1}{x+2}\right)=\frac{\pi}{12} \quad\left[\because \cot ^{-1} x=\frac{\tan ^{-1} 1}{x}\right]\\
&\Rightarrow \tan ^{-1}\left(\frac{\frac{1}{x}-\frac{1}{x+2}}{1+\frac{1}{x(x+2)}}\right)=\frac{\pi}{12}\\
&\Rightarrow \tan ^{-1}\left(\frac{\frac{2}{x(x+2)}}{\frac{x^{2}+2 x+1}{x(x+2)}}\right)=\frac{\pi}{12}\\
&\Rightarrow \tan ^{-1}\left(\frac{2}{x^{2}+2 x+1}\right)=\frac{\pi}{12}\\
&\Rightarrow\left(\frac{2}{x^{2}+2 x+1}\right)=\tan \frac{\pi}{12}\\
&\Rightarrow\left(\frac{2}{x^{2}+2 x+1}\right)=\tan \left(\frac{\pi}{3}-\frac{\pi}{4}\right)\\
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