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

 

Dear Student,
Please find below the solution to your problem.

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

Thanks and Regards