If the sum of all the solutions of tan - 1 ( 2 x 1 - x 2 ) + cot - 1 ( 1 - x 2 2 x ) = 3 , - 1 x 1 , x 0 , is 4 3 , then is equal to _______ . [2023]

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Solution

Q.
If the sum of all the solutions of $\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{π}{3}, - 1 < x < 1, x \neq 0,$ is $α - \frac{4}{\sqrt{3}},$ then $α$ is equal to _______ . [2023]

Ans.
(2)

$Given, \tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{π}{3}$

$Case I : If x > 0$

$\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{π}{3}$ $\Rightarrow \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{π}{6}$

$\Rightarrow \frac{2 x}{1 - x^{2}} = \frac{1}{\sqrt{3}} \Rightarrow x^{2} + 2 \sqrt{3} x - 1 = 0$

$\Rightarrow [x - (2 - \sqrt{3})] [x + (2 + \sqrt{3})] = 0 \Rightarrow x = 2 - \sqrt{3}, - (2 + \sqrt{3})$

$x = - (2 + \sqrt{3}) is rejected because x > 0 . Hence, x = 2 - \sqrt{3}$

$Case II : If x < 0$

$\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{π}{3} - π = - \frac{2 π}{3}$

$\Rightarrow \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = - \frac{π}{3} \Rightarrow \frac{2 x}{1 - x^{2}} = - \sqrt{3}$

$\Rightarrow \sqrt{3} x^{2} - 2 x - \sqrt{3} = 0 \Rightarrow (\sqrt{3} x - 3) (x + \frac{1}{\sqrt{3}}) = 0$

$\Rightarrow x = \frac{- 1}{\sqrt{3}}, \sqrt{3} \Rightarrow x = \sqrt{3} is rejected because x < 0 .$

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

$∴$ $Sum of solutions = 2 - \sqrt{3} - \frac{1}{\sqrt{3}} = 2 - \frac{4}{\sqrt{3}}$

$α - \frac{4}{\sqrt{3}} = 2 - \frac{4}{\sqrt{3}} \Rightarrow α = 2$

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