Question

cos^-1 ( (1-x^2) / (1+x^2)) + tan^-1 (2x/(x^2-1)) = 2pi/3

Answer 1

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

Answer 2

You may wish to move the arctan to the other side, apply the phantom cosine, and expand the new right hand side with the difference of cosines.

You may have to come to terms with the sign of things.

\(1 + x^{2} > 0\),

But \(1-x^{2}\) has stipulations and is always a different sign (except for \(x=1\) than \(x^{2}-1\)