JEE Main 2026 (Online) 4th April Morning ShiftQuestion 1
Let $\vec a_k=(\tan\theta_k)\hat i+\hat j$ and $\vec b_k=\hat i-(\cot\theta_k)\hat j$, where $\theta_k=\frac{2^{k-1}\pi}{2^{n}+1}$, for some $n\in\mathbb{N}$, $n>5$. Then the value of $\frac{\sum_{k=1}^{n}|\vec a_k|^{2}}{\sum_{k=1}^{n}|\vec b_k|^{2}}$ is ________.
Vector Algebra
Answer$3$
Solution
$|\vec a_k|^2=\sec^2\theta_k$, $|\vec b_k|^2=\csc^2\theta_k$. Since $\cot\theta_k-\tan\theta_k=2\cot2\theta_k=2\cot\theta_{k+1}$ and $\theta_{n+1}=\pi-\theta_1$, squaring and summing gives $T+C-2n=4C$, i.e. $T+n=3(C+n)$. Ratio $=3$.