For x∈R, $$tan$$⁡$$x+12tan$$⁡$$x2+122tan$$⁡$$x22+$$…$$+12n$$−$$1tan$$⁡$$x2n$$−$$1$$ is equal to

Question

For x∈R, $$tan$$⁡$$x+12tan$$⁡$$x2+122tan$$⁡$$x22+$$…$$+12n$$−$$1tan$$⁡$$x2n$$−$$1$$ is equal to

Infinity Learn · Accepted Answer

We have,$$tan$$⁡$$x=$$$$cot$$⁡x−$$2cot$$⁡$$2x$$⇒$$12tan$$⁡$$x2=12cot$$⁡$$x2$$−$$cot$$⁡x⇒$$12tan$$⁡$$x22=12cot$$⁡$$x22$$−$$cot$$⁡$$x2$$⇒$$122tan$$⁡$$x22=122cot$$⁡$$x22$$−$$12cot$$⁡$$x2Similarly$$, we $$have123tan$$⁡$$x23=123cot$$⁡$$x23$$−$$122cot$$⁡$$x22$$........................................................  .......................................................$$.12n$$−$$1tan$$⁡$$x2n$$−$$1=12n$$−$$1cot$$⁡$$x2n$$−$$1$$−$$12n$$−$$2cot$$⁡$$x2n$$−$$2Adding$$ a II the above results, we gettan⁡$$x+12tan$$⁡$$x2+122tan$$⁡$$x22+$$…$$+12n$$−$$1tan$$⁡$$x2n$$−$$1=12n$$−$$1cot$$⁡$$x2n$$−$$1$$−$$2cot$$⁡$$2x$$

For $x \in R$ , $\tan x + \frac{1}{2} \tan \frac{x}{2} + \frac{1}{2^{2}} \tan \frac{x}{2^{2}} + \dots + \frac{1}{2^{n - 1}} \tan (\frac{x}{2^{n - 1}})$ is equal to

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For x∈R, tan⁡x+12tan⁡x2+122tan⁡x22+…+12n−1tan⁡x2n−1 is equal to

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For $x \in R$ , $\tan x + \frac{1}{2} \tan \frac{x}{2} + \frac{1}{2^{2}} \tan \frac{x}{2^{2}} + \dots + \frac{1}{2^{n - 1}} \tan (\frac{x}{2^{n - 1}})$ is equal to