JEE Main 2022 · 29 Jun · Shift 1, Mathematics question 26

First term. With $\tan^{-1}\frac12=a$, $\cos^{-1}\frac1{\sqrt5}=\tan^{-1}2=\frac\pi2-a$, so the angle is $\pi+a$ and $\tan=\frac12$: $50\cdot\frac12=25$.

Second term. $\tan\frac\phi2$ with $\tan\phi=2\sqrt2$, $\cos\phi=\frac13$: $\tan\frac\phi2=\frac1{\sqrt2}$, giving $4\sqrt2\cdot\frac1{\sqrt2}=4$.

Total 29.

Answer: 29