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Prove that : Sec^6x-tan^6x= 1+3tan^2x+ 3tan^4x

 

Asked by siddharthlathia0 10th June 2018, 6:55 AM
Answered by Expert
Answer:
Check your question

a³-b³=(a-b)³+3ab(a-b)
1+tan²x=sec²x
(sec
²x)³-(tan²x)³=(sec²x-tan²x)³+3sec²xtan²x(sec²x-tan²x)

                           =1³+3sec²xtan2x(1)
                             =1+3tan²xsec²x
Answered by Expert 11th June 2018, 11:07 AM
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