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

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Inverse Trigonometric Functions

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}{3}

, then

α

is equal to ______.

Answer: 2

For

x > 0

both arguments are positive, so

cot^{- 1} \frac{1 - x ^{2}}{2 x} = tan^{- 1} \frac{2 x}{1 - x ^{2}}

and the equation is

2 tan^{- 1} \frac{2 x}{1 - x ^{2}} = \frac{π}{3}

, giving

\frac{2 x}{1 - x ^{2}} = \frac{1}{3}

and

x = 2 - 3

. For

x < 0

the cotangent term carries an extra

π

, leading to

\frac{2 x}{1 - x ^{2}} = - 3

and

x = - \frac{1}{3}

. The sum of solutions is

(2 - 3) - \frac{1}{3} = 2 - \frac{4}{3}

, so

α = 2

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