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

							
Dear Paradox
Solving from right
tan 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/tan4A

Proceeding as above, we will reach to
tan A + 2tan2A + 4tan4A + 8cot8A = 1/tanA = cotA
Hence proved.
All the best.
AKASH GOYAL
AskiitiansExpert-IITD
 
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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