Prove that tan⁻¹1/2 = π/4 - 1/2 cos⁻¹(4/5).

Prove that $\tan^{-1}\frac{1}{2} = \frac{\pi}{4} - \frac{1}{2} \cos^{-1}\left(\frac{4}{5}\right)$.

To show: $\tan^{-1}\frac{1}{2} = \frac{\pi}{4} - \frac{1}{2} \cos^{-1}\left(\frac{4}{5}\right)$

Answer

Proof that $\tan^{-1}\frac{1}{2} = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}\frac{4}{5}$.

Let $\alpha = \cos^{-1}\frac{4}{5}$, so $\cos\alpha = \frac{4}{5}$ and $0 < \alpha < \frac{\pi}{2}$.
$\tan^2\frac{\alpha}{2} = \frac{1 - \cos\alpha}{1 + \cos\alpha} = \frac{1/5}{9/5} = \frac{1}{9}$ (half-angle identity).
Since $\frac{\alpha}{2}$ is acute, $\tan\frac{\alpha}{2} = \frac{1}{3}$, so $\frac{1}{2}\cos^{-1}\frac{4}{5} = \tan^{-1}\frac{1}{3}$.
RHS $= \frac{\pi}{4} - \tan^{-1}\frac{1}{3} = \tan^{-1}1 - \tan^{-1}\frac{1}{3}$.
Using $\tan^{-1}a - \tan^{-1}b = \tan^{-1}\frac{a - b}{1 + ab}$ (here $ab > -1$): RHS $= \tan^{-1}\frac{1 - 1/3}{1 + 1/3}$.
$= \tan^{-1}\frac{2/3}{4/3} = \tan^{-1}\frac{1}{2} = $ LHS.
Hence $\tan^{-1}\frac{1}{2} = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}\left(\frac{4}{5}\right)$.