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

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

Given: $\frac{\pi}{4} - \frac{1}{2}\cos^{-1}\left(\frac{4}{5}\right)$

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

Answer

Let $\cos^{-1}\frac{4}{5} = \theta$, so $\cos\theta = \frac{4}{5}$, $0 < \theta < \frac{\pi}{2}$. $\tan\frac{\theta}{2} = \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} = \sqrt{\frac{1/5}{9/5}} = \frac{1}{3}$. $\tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = \frac{1 - \frac{1}{3}}{1 + \frac{1}{3}} = \frac{1}{2}$. As $0 < \frac{\pi}{4} - \frac{\theta}{2} < \frac{\pi}{4}$ lies in the principal range, $\frac{\pi}{4} - \frac{\theta}{2} = \tan^{-1}\frac{1}{2}$. Hence $\tan^{-1}\frac{1}{2} = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}\frac{4}{5}$.

Let $\theta = \cos^{-1}\frac{4}{5}$, so $\cos\theta = \frac{4}{5}$ with $\theta \in (0, \frac{\pi}{2})$.
$\tan\frac{\theta}{2} = \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} = \frac{1}{3}$, by the half-angle formula.
$\tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = \frac{1 - \tan\frac{\theta}{2}}{1 + \tan\frac{\theta}{2}} = \frac{2/3}{4/3} = \frac{1}{2}$.
$\frac{\pi}{4} - \frac{\theta}{2} \in (0, \frac{\pi}{4})$, within the principal range of $\tan^{-1}$.
Hence $\tan^{-1}\frac{1}{2} = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}\frac{4}{5}$.