I was studying trigonometry and I found a good prove problem which is

**Prove that:**\(\tan { A } +2\tan { 2A } +4\tan { 4A } +8\cot { 8A } =\cot { A } \).

I have a non-bashing solution,want to see your approach.

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TopNewestTake tan on RHS and use the property

\(\displaystyle cotA - tanA = 2cot(2A)\)

Again take 2tan2A on RHS

\(2cot2A - 2tan2A = 4cot4A\)

Again bring 4tan4A on RHS

\(4cot4A - 4tan4A = 8cot8A\) on RHS

I think there is a typo it should be 8cot8A instead of 8tan8A on LHS.

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Ya I was talking of same non-bashing solution.Is there any shorter or equivalent method??

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That's the shortest and simplest solution according to me, I cannot think of any 'shorter' solution right now.

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