Question

cos^-1.[(x^2)-1]/[(x^2)+1] + tan^-1.[2x]/[(x^2)-1] = 2(pi)/3
solve for x

Answer 1

\[\cos^{-1}\left ( \frac{x^2-1}{x^2+1}\right )+\tan^{-1} \left ( \frac{2x}{x^2-1}\right )=\frac{2\pi}{3}\]
Let \(\Large\theta=\tan^{-1} \left ( \frac{2x}{x^2-1}\right ) \Rightarrow \tan\theta=\frac{2x}{x^2-1}\).
Now draw a right triangle with this theta in it and write the arccos in terms of theta.

Answer 2

using the substitution @blockcolder used, the question reduce to 2(theeta)=2(pi)/3.
then i think the question is straight forward