Solve cos⁻¹((x²−1)/(x²+1)) + ½tan⁻¹(2x/(1−x²))

Solve \( \cos^{-1}\left(\frac{x^2 – 1}{x^2 + 1}\right) + \frac{1}{2}\tan^{-1}\left(\frac{2x}{1 – x^2}\right) = \frac{2\pi}{3} \)

Solution:

Use identities:

\[ \cos^{-1}\left(\frac{1 – t^2}{1 + t^2}\right) = 2\tan^{-1}(t) \]

Here,

\[ \frac{x^2 – 1}{x^2 + 1} = -\frac{1 – x^2}{1 + x^2} \]

So,

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

Also,

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

Substitute:

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

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

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

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

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

\[ x = \sqrt{3} \]

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