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```				   prove that
tanA + 2tan2A + 4tan4A + 8cot8A = cotA
```

5 years ago

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```										Dear Paradox
Solving from righttan A + 2tan2A + 4tan4A + 8cot8A = tan A + 2tan2A + 4tan4A + 8/tan8A = tan A + 2tan2A + 4tan4A + 8(1-tan24A)/2tan4A = tan A + 2tan2A + [{4tan4A(tan4A)} + 4 (1-tan24A)]/tan4A = tan A + 2tan2A + [4tan24A + 4 - tan24A]/tan4A   = tan A + 2tan2A + 4/tan4AProceeding as above, we will reach totan A + 2tan2A + 4tan4A + 8cot8A = 1/tanA = cotAHence proved.
All the best.
AKASH GOYAL

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```
5 years ago
```										TANA + 2TAN2A + 4TAN4A + 8COT8A = COTA .........1
WE HAVE FORMULA , COTA-TANA=2COT2A

EQ 1 CAN BE WRITTEN AS
2TAN2A + 4TAN4A + 8COT8A = COTA-TANA
2TAN2A + 4TAN4A + 8COT8A = 2COT2A               ( BY USING FORMULA )
4TAN4A + 8COT8A = 2COT2A - 2TAN2A
4TAN4A + 8COT8A =4COT4A                               (BY USING FORMULA)
8COT8A = 4COT4A - 4TAN4A
=8COT8A                        (BY USING FORMULA )
HENCE PROVED

```
5 years ago

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