MathsJEE Main 4 Apr 2026, Shift 1·Q24·Numerical·Tough

Trigonometry · Trigonometric sum with doubling angles

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 = \dfrac{2^{k-1}\pi}{2^n + 1}

, for some

n \in \mathbb{N}

n > 5

. Then the value of

\dfrac{\displaystyle\sum_{k=1}^{n}\lvert\vec a_k\rvert^2}{\displaystyle\sum_{k=1}^{n}\lvert\vec b_k\rvert^2}

is .

We corrected a printing error in the original paper

Source printed “ $\vec b_k = \hat i - (\cos\theta_k)\hat j$ ”; corrected to the intended “ $(\cot\theta_k)\hat j$ ” (with $\cos$ the ratio depends on $n$ , e.g. 166.5 for $n = 6$ , and is never 3; with $\cot$ it is 3 for every $n$ , as the key and the source's solution use).

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See the full 4 Apr 2026, Shift 1 paper