JEE Main 2026 — 4 April, Morning Shift — Question 46

$\left|\mathrm{a}_{\mathrm{k}}\right|^{2}=1+\tan ^{2} \theta_{\mathrm{k}}=\sec ^{2} \theta_{\mathrm{k}}$ $\left|b_{k}\right|^{2}=1+\cot ^{2} \theta_{k}=\operatorname{cosec}^{2} \theta_{k}$ Since, $\cot \theta-\tan \theta=2 \cot 2 \theta$ $\therefore-\operatorname{cosec}^{2} \theta-\sec ^{2} \theta=-4 \operatorname{cosec}^{2} 2 \theta$ $\sec ^{2} \theta=4 \operatorname{cosec}^{2} 2 \theta-\operatorname{cosec}^{2} \theta$ $\Sigma \sec ^{2} \theta_{\mathrm{k}}=4 \Sigma \operatorname{cosec}^{2} 2 \theta_{\mathrm{k}}-\Sigma \operatorname{cosec}^{2} \theta_{\mathrm{k}}$ Now $\Sigma \operatorname{cosec}^{2} 2 \theta_{\mathrm{k}}=\Sigma \operatorname{cosec}^{2} \theta_{\mathrm{k}}$ since $\sum_{\mathrm{k}=1}^{\mathrm{n}} \operatorname{cosec}^{2} \frac{2^{\mathrm{k}} \pi}{2^{\mathrm{n}}+1}=\sum_{\mathrm{k}=1}^{\mathrm{n}} \operatorname{cosec}^{2} \frac{2^{\mathrm{k}-1} \pi}{2^{\mathrm{n}}+1}$ because $\operatorname{cosec} \frac{2^{n} \pi}{2^{n}+1}=\operatorname{cosec} \frac{\pi}{2^{n}+1}$ $\therefore \Sigma \sec ^{2} \theta_{\mathrm{k}}=3 \Sigma \operatorname{cosec}^{2} \theta_{\mathrm{k}}$ $\frac{\sum \sec ^{2} \theta_{\mathrm{k}}}{\sum \operatorname{cosec}^{2} \theta_{\mathrm{k}}}=3$