Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Given,

${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+{\cot }^{-1}\left(\frac{1-{\mathrm{x}}^2}{2\mathrm{x}}\right)=\frac{\mathrm{\pi }}{3}$ where $x\in \left(-1,1\right)$

Case $1$: When $x<0$, then

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

$\Rightarrow 2{\tan }^{-1}x+\mathrm{\pi }+2{\tan }^{-1}\mathrm{x}=\frac{\mathrm{\pi }}{3}$

$\Rightarrow 4{\tan }^{-1}x=-\mathrm{\pi }+\frac{\mathrm{\pi }}{3}$

$\Rightarrow 4{\tan }^{-1}x=-\frac{2\mathrm{\pi }}{3}$

$\Rightarrow {\tan }^{-1}x=-\frac{\mathrm{\pi }}{6}$

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

Case $2$: When $x\ge 0$,

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

$\Rightarrow 2{\tan }^{-1}x+2{\tan }^{-1}x=\frac{\mathrm{\pi }}{3}$

$\Rightarrow {\tan }^{-1}x=\frac{\mathrm{\pi }}{12}$

$\Rightarrow x=2-\sqrt{3}$

So, sum of solution will be 

$\frac{-1}{\sqrt{3}}+2-\sqrt{3}=2-\frac{4}{\sqrt{3}}$

Now, on comparing with $\alpha -\frac{4}{\sqrt{3}}$, we get

$\alpha =2$