Solve tan⁻¹(2x/(1−x²)) + cot⁻¹((1−x²)/2x) = 2π/3

Solve \( \tan^{-1}\left(\frac{2x}{1-x^2}\right) + \cot^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{2\pi}{3}, \quad x > 0 \)

Solution:

Use identity:

\[ \cot^{-1}(t) = \tan^{-1}\left(\frac{1}{t}\right) \]

So,

\[ \cot^{-1}\left(\frac{1-x^2}{2x}\right) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \]

Thus equation becomes:

\[ 2\tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{2\pi}{3} \]

\[ \Rightarrow \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{\pi}{3} \]

Taking tangent:

\[ \frac{2x}{1-x^2} = \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]

\[ 2x = \sqrt{3}(1 – x^2) \]

\[ \sqrt{3}x^2 + 2x – \sqrt{3} = 0 \]

Solve quadratic:

\[ x = \frac{-2 \pm \sqrt{4 + 12}}{2\sqrt{3}} = \frac{-2 \pm 4}{2\sqrt{3}} \]

\[ x = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}, \quad x = -\sqrt{3} \]

Since \(x > 0\),

\[ x = \frac{1}{\sqrt{3}} \]

\[ x = \frac{1}{\sqrt{3}} \]

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