Question

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 ___

Answer 1

Correct Answer: 2

The correct answer is 2.
Case $I:x>0$
${\tan }^{-1}\frac{2x}{1-x^2}+{\tan }^{-1}\frac{2x}{1-x^2}=\frac{\pi }{3}$
$x=2-\sqrt{3}$
Case II : $x<0$
${\tan }^{-1}\frac{2x}{1-x^2}+{\tan }^{-1}\frac{2x}{1-x^2}+\pi =\frac{\pi }{3}$
$x=\frac{-1}{\sqrt{3}}\Rightarrow \alpha =2$

Answer 2

1. Simplify the Equation: Use the properties of inverse trigonometric functions: \[ \tan^{-1}\left(\frac{2x}{1-x^2}\right) + \cot^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{\pi}{2}. \] Simplify: \[ \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{\pi}{6}. \] 2. Find \( x \): Using \( \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \): \[ \frac{2x}{1-x^2} = \frac{1}{\sqrt{3}}. \] Cross-multiply: \[ 2\sqrt{3}x = 1 - x^2 \implies x^2 + 2\sqrt{3}x - 1 = 0. \] 3. Solve the Quadratic Equation: Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{-2\sqrt{3} \pm \sqrt{12 + 4}}{2} = \frac{-2\sqrt{3} \pm \sqrt{16}}{2}. \] \[ x = -\sqrt{3} \quad \text{or} \quad x = \frac{-\sqrt{3}}{2}. \] 4. Sum of Solutions: The sum of solutions is: \[ \alpha - \frac{4}{\sqrt{3}} = 2, \quad \alpha = 2. \]