JEE Main 2026 — 5 April, Morning Shift — Question 47

$\alpha=\frac{\pi}{4}+\sum_{p=1}^{11} \tan ^{-1}\left(\frac{2^{p-1}}{1+2^{2 p-1}}\right)$ $=\frac{\pi}{4}+\sum_{p=1}^{11} \tan ^{-1}\left(\frac{2^{p}-2^{p-1}}{1+2^{p} 2^{p-1}}\right)$ $=\frac{\pi}{4}+\sum_{p=1}^{11}\left(\tan ^{-1}\left(2^{p}\right)-\tan ^{-1}\left(2^{p-1}\right)\right)$ $=\frac{\pi}{4}+\tan ^{-1}\left(2^{11}\right)-\tan ^{-1}\left(2^{0}\right)=2^{11}=2048$