Vectors & 3D GeometrymediumPYQ · JEE Main · 4 Apr 2026 · Shift 1 (Morning)Free

Sum of sec² over cosec² Telescoping: 3 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let a⃗k=(tan⁡θk)i^+j^\vec a_k=(\tan\theta_k)\hat i+\hat j and b⃗k=i^−(cot⁡θk)j^\vec b_k=\hat i-(\cot\theta_k)\hat j, where θk=2k−1π2n+1\theta_k=\dfrac{2^{k-1}\pi}{2^n+1}, for some n∈N, n>5n\in\mathbb{N},\ n>5. Then the value of ∑k=1n∣a⃗k∣2∑k=1n∣b⃗k∣2\dfrac{\displaystyle\sum_{k=1}^{n}|\vec a_k|^2}{\displaystyle\sum_{k=1}^{n}|\vec b_k|^2} is
Solution
Answer: 3 (± 0.01)
Step 1: ∣a⃗k∣2=1+tan⁡2θk=sec⁡2θk|\vec a_k|^2=1+\tan^2\theta_k=\sec^2\theta_k and ∣b⃗k∣2=1+cot⁡2θk=cosec⁡2θk|\vec b_k|^2=1+\cot^2\theta_k=\operatorname{cosec}^2\theta_k. Step 2: Use cot⁡θ−tan⁡θ=2cot⁡2θ\cot\theta-\tan\theta=2\cot2\theta, which gives sec⁡2θ=4cosec⁡22θ−cosec⁡2θ\sec^2\theta=4\operatorname{cosec}^2 2\theta-\operatorname{cosec}^2\theta. Summing the telescoping part,
∑sec⁡2θk=4∑cosec⁡22θk−∑cosec⁡2θk.\sum \sec^2\theta_k=4\sum\operatorname{cosec}^2 2\theta_k-\sum\operatorname{cosec}^2\theta_k.
 Step 3: With θk=2k−1π2n+1\theta_k=\dfrac{2^{k-1}\pi}{2^n+1}, the angles satisfy cosec⁡22nπ2n+1=cosec⁡2π2n+1\operatorname{cosec}^2\dfrac{2^n\pi}{2^n+1}=\operatorname{cosec}^2\dfrac{\pi}{2^n+1}, so ∑cosec⁡22θk=∑cosec⁡2θk\sum\operatorname{cosec}^2 2\theta_k=\sum\operatorname{cosec}^2\theta_k. Therefore
∑sec⁡2θk=4∑cosec⁡2θk−∑cosec⁡2θk=3∑cosec⁡2θk.\sum\sec^2\theta_k=4\sum\operatorname{cosec}^2\theta_k-\sum\operatorname{cosec}^2\theta_k=3\sum\operatorname{cosec}^2\theta_k.
 Step 4:
∑sec⁡2θk∑cosec⁡2θk=3.\frac{\sum\sec^2\theta_k}{\sum\operatorname{cosec}^2\theta_k}=3.
 Correct answer: 3
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