If the sum of all the solutions of tan ^{-1}l | vedclass

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(B) Given equation: $	an ^{-1}\left(\frac{2 x}{1-x^2}ight)+\cot ^{-1}\left(\frac{1-x^2}{2 x}ight)=\frac{\pi}{3}$.Since $\cot ^{-1}(y) = 	an ^{-1

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$2$

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(B) Given equation: $	an ^{-1}\left(\frac{2 x}{1-x^2}ight)+\cot ^{-1}\left(\frac{1-x^2}{2 x}ight)=\frac{\pi}{3}$.Since $\cot ^{-1}(y) = 	an ^{-1}(\frac{1}{y})$ for $y > 0$,we have $\cot ^{-1}(\frac{1-x^2}{2x}) = 	an ^{-1}(\frac{2x}{1-x^2})$ for $\frac{1-x^2}{2x} > 0$,i.e.,$x(1-x^2) > 0$.Case $I$: $x(1-x^2) > 0$. This holds for $x \in (-1, 0) \cup (0, 1)$.If $x \in (0, 1)$,then $\frac{2x}{1-x^2} > 0$,so $	an ^{-1}(\frac{2x}{1-x^2}) + 	an ^{-1}(\frac{2x}{1-x^2}) = \frac{\pi}{3} \Rightarrow 2 	an ^{-1}(\frac{2x}{1-x^2}) = \frac{\pi}{3} \Rightarrow 	an ^{-1}(\frac{2x}{1-x^2}) = \frac{\pi}{6}$.Then $\frac{2x}{1-x^2} = 	an(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} \Rightarrow x^2 + 2\sqrt{3}x - 1 = 0$. Solving for $x > 0$,$x = \frac{-2\sqrt{3} + \sqrt{12+4}}{2} = 2 - \sqrt{3}$.Case $II$: $x(1-x^2) Then $\cot ^{-1}(\frac{1-x^2}{2x}) = \pi + 	an ^{-1}(\frac{2x}{1-x^2})$.The equation becomes $2 	an ^{-1}(\frac{2x}{1-x^2}) + \pi = \frac{\pi}{3} \Rightarrow 2 	an ^{-1}(\frac{2x}{1-x^2}) = -\frac{2\pi}{3} \Rightarrow 	an ^{-1}(\frac{2x}{1-x^2}) = -\frac{\pi}{3}$.Then $\frac{2x}{1-x^2} = 	an(-\frac{\pi}{3}) = -\sqrt{3} \Rightarrow \sqrt{3}x^2 - 2x - \sqrt{3} = 0$. Solving for $x The sum of solutions is $(2 - \sqrt{3}) + (-\frac{1}{\sqrt{3}}) = 2 - \frac{3+1}{\sqrt{3}} = 2 - \frac{4}{\sqrt{3}}$.Comparing with $\alpha - \frac{4}{\sqrt{3}}$,we get $\alpha = 2$.

If the sum of all the solutions of $\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)+\cot ^{-1}\left(\frac{1-x^2}{2 x}\right)=\frac{\pi}{3}$ for $-1 < x < 1$ and $x \neq 0$ is $\alpha-\frac{4}{\sqrt{3}}$,then $\alpha$ is equal to $..........$.

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