Evaluate cos⁻¹(1 - x²)/(1 + x²) at x = -1/2…

Put $x = \tan t$ for $1 + x^2$, and $x = \sin t$ for $\sqrt{1 - x^2}$. Then $\tan^{-1}\dfrac{\sqrt{1 + x^2} - 1}{x} = \dfrac{1}{2}\tan^{-1}x$, and $\cos^{-1}\dfrac{1 - x^2}{1 + x^2} = 2\left|\tan^{-1}x\right|$. The fold rules: $\sin^{-1}(3x - 4x^3) = 3\sin^{-1}x$ for $|x| \le \dfrac{1}{2}$, and for $x > \dfrac{1}{2}$ it is $\pi - 3\sin^{-1}x$.