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

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