let an=∑k=1n tan2kπ2n+1& bn=∏k=1n tan2kπ2n+1, then

⇒sin⁡nθ=∑k=0n−12 (−1)knC2k+1cosn−2k−1⁡θsin(2k+1)⁡θsin⁡(2n+1)θ=∑k=0n (−1)k2n+1C2k+1cos2n−2k⁡θsin2k+1⁡θ=tan⁡θcos2n+1⁡θ∑k=0n (−1)k⋅2n+1C2k+1(tan⁡θ)2k∑k=0n (−1)k2n+1C2k+1tan2k⁡θ=0 for θ=jπ2n+1,1≤j≤n so tan2⁡jπ2n+1,1≤j≤nare the roots of ∑k=0n (−1)k⋅2n+1C2k+1xk=0∵an&bnare the sum and the product of the roots respectively, we havean=2n+1C2=n(2n+1)bn=2n+1C2n=2n+1

let an=∑k=1n tan2kπ2n+1& bn=∏k=1n tan2kπ2n+1, then

⇒sin⁡nθ=∑k=0n−12 (−1)knC2k+1cosn−2k−1⁡θsin(2k+1)⁡θsin⁡(2n+1)θ=∑k=0n (−1)k2n+1C2k+1cos2n−2k⁡θsin2k+1⁡θ=tan⁡θcos2n+1⁡θ∑k=0n (−1)k⋅2n+1C2k+1(tan⁡θ)2k∑k=0n (−1)k2n+1C2k+1tan2k⁡θ=0 for θ=jπ2n+1,1≤j≤n so tan2⁡jπ2n+1,1≤j≤nare the roots of ∑k=0n (−1)k⋅2n+1C2k+1xk=0∵an&bnare the sum and the product of the roots respectively, we havean=2n+1C2=n(2n+1)bn=2n+1C2n=2n+1

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let $a_{n} = \sum_{k = 1}^{n} \tan^{2} \frac{k π}{2 n + 1} & b_{n} = \prod_{k = 1}^{n} \tan^{2} \frac{k π}{2 n + 1}$ , then

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Units digit in $b_{2022}$ is five

Units digit in $b_{2022}$ is three

Units digit in $a_{2022}$ is zero

Units digit in $a_{2022}$ is nine

answer is A, C.

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Detailed Solution

$\Rightarrow \sin n θ = \sum_{k = 0}^{|\frac{n - 1}{2}|} (- 1)^{k n} C_{2 k + 1} \cos^{n - 2 k - 1} θ \sin^{(2 k + 1)} θ$
$\sin (2 n + 1) θ = \sum_{k = 0}^{n} {(- 1)^{k}}^{2 n + 1} C_{2 k + 1} \cos^{2 n - 2 k} θ \sin^{2 k + 1} θ$
$= \tan θ \cos^{2 n + 1} θ \sum_{k = 0}^{n} (- 1)^{k} \cdot^{2 n + 1} C_{2 k + 1} (\tan θ)^{2 k}$
$\sum_{k = 0}^{n} (- 1)^{k 2 n + 1} C_{2 k + 1} \tan^{2 k} θ = 0$
$for θ = \frac{j π}{2 n + 1}, 1 \leq j \leq n$
$so \tan^{2} \frac{j π}{2 n + 1}, 1 \leq j \leq n$ are the roots of $\sum_{k = 0}^{n} (- 1)^{k} \cdot^{2 n + 1} C_{2 k + 1} x^{k} = 0$
$∵ a_{n} & b_{n}$ are the sum and the product of the roots respectively, we have
$\begin{aligned} a_{n} =^{2 n + 1} C_{2} = n (2 n + 1) \\ b_{n} =^{2 n + 1} C_{2 n} = 2 n + 1 \end{aligned}$