JEE Advanced 2022 · Paper 1, Mathematics question 1

One angle. Let $\theta=\cos^{-1}\sqrt{\dfrac{2}{2+\pi^2}}$, with $\theta\in(0,\frac\pi2)$. Then
$$\cos\theta=\frac{\sqrt2}{\sqrt{2+\pi^2}},\quad\sin\theta=\frac{\pi}{\sqrt{2+\pi^2}},\quad\tan\theta=\frac{\pi}{\sqrt2}\approx2.22,$$
so $\theta\approx1.148>\frac\pi4$.

The other two terms.

• $\dfrac{2\sqrt2\,\pi}{2+\pi^2}=2\sin\theta\cos\theta=\sin2\theta$. Since $2\theta\in(\frac\pi2,\pi)$, the principal value gives $\sin^{-1}(\sin2\theta)=\pi-2\theta$.
• $\tan^{-1}\dfrac{\sqrt2}{\pi}=\tan^{-1}(\cot\theta)=\dfrac\pi2-\theta$.

Sum.
$$\frac32\theta+\frac14(\pi-2\theta)+\frac\pi2-\theta=\frac{3\pi}4\approx2.36.$$