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Miscellaneous Exercise 3 (II) · Q106

Q.Prove the following: tan⁡A+2tan⁡2A+4tan⁡4A+8cot⁡8A=cot⁡A\tan A+2\tan2A+4\tan4A+8\cot8A=\cot A

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Step 1: From cot⁡θ−tan⁡θ=2cot⁡2θ\cot\theta-\tan\theta=2\cot2\theta: tan⁡θ=cot⁡θ−2cot⁡2θ\tan\theta=\cot\theta-2\cot2\theta.

Step 2: At θ=A\theta=A: tan⁡A=cot⁡A−2cot⁡2A\tan A=\cot A-2\cot2A. At θ=2A\theta=2A: 2tan⁡2A=2cot⁡2A−4cot⁡4A2\tan2A=2\cot2A-4\cot4A. At θ=4A\theta=4A: 4tan⁡4A=4cot⁡4A−8cot⁡8A4\tan4A=4\cot4A-8\cot8A. …

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