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$\tan x + \frac{1}{2} \tan \frac{x}{2} + \frac{1}{2^{2}} \tan \frac{x}{2^{2}} + \dots + \frac{1}{2^{n - 1}} \tan \frac{x}{2^{n - 1}}$ is equal to

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a

12ncot⁡x2n−2cot⁡2x

b

12n−1cot⁡x2n−1−2cot⁡2x

c

tan⁡2n−12n−1x

d

2cot⁡2x−12n−1cot⁡x2n−1

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detailed solution

Correct option is B

We can write tan x=tan2⁡x−1+1tan⁡x=cot⁡x−2cot⁡2x so tan⁡x+12tan⁡x2+122tan⁡x22+⋯+12n−1tan⁡x2n−1 =cot⁡x−2cot⁡2x+12cot⁡x2−2cot⁡x+122cot⁡x22−2cot⁡x2+⋯+12n−1cot⁡x2n−1−2cot⁡x2n−2 =12n−1cot⁡x2n−1−2cot⁡ 2x

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