Let \(\vec{a}_{k}=\left(\tan \theta_{k}\right) \hat{i}+\hat{j}\) and \(\vec{b}_{k}=\hat{i}-\left(\cos \theta_{k}\right) \hat{j}\), where \(\theta_{k}=\frac{2^{k-1} \pi}{2^{n}+1}\), for some \(n \in \mathbb{N}, n>5\). Then the value of \(\frac{\sum_{k=1}^{n}\left|\vec{a}_{k}\right|^{2}}{\sum_{k=1}^{n}\left|\vec{b}_{k}\right|^{2}}\) is ____.

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Vector Algebra

Let

a_{k} = (tan θ_{k}) \hat{i} + \hat{j}

and

b_{k} = \hat{i} - (cos θ_{k}) \hat{j}

, where

θ_{k} = \frac{2 ^{k - 1} π}{2 ^{n} + 1}

, for some

n \in N, n > 5

. Then the value of

\frac{\sum _{k = 1}^{n} ∣ a _{k} ∣ ^{2}}{\sum _{k = 1}^{n} ∣ b _{k} ∣ ^{2}}

is ____.

Answer: 3

With

∣ a_{k} ∣^{2} = tan^{2} θ_{k} + 1 = sec^{2} θ_{k}

and

∣ b_{k} ∣^{2} = 1 + cot^{2} θ_{k} = csc^{2} θ_{k}

, start from

cot θ - tan θ = 2 cot 2 θ

, which on squaring and rearranging gives

sec^{2} θ = 4 csc^{2} 2 θ - csc^{2} θ

. Because

2 θ_{k} = θ_{k + 1}

and

θ_{n + 1} = π - θ_{1}

, the doubled angles satisfy

k = 1 \sum n csc^{2} 2 θ_{k} = k = 1 \sum n csc^{2} θ_{k}

. Hence

\sum sec^{2} θ_{k} = 4 \sum csc^{2} θ_{k} - \sum csc^{2} θ_{k} = 3 \sum csc^{2} θ_{k}

, so the required ratio

\frac{\sum ∣ a _{k} ∣ ^{2}}{\sum ∣ b _{k} ∣ ^{2}} = 3

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Integer answer