Question

sec ^ 6 x-tan ^ 6 x = 1 + 3 tan^2 + 3 tan ^ 4 how to do it please tell​

Answer 1

Recall that 1 + tan2x = sec2x  and A3-B3 = (A-B)(A2+AB+B2)

 

sec6x - tan6x = (sec2x)3 - (tan2x)3

 

                   = (sec2x - tan2x)(sec4x + sec2xtan2x + tan4x)

 

                   = (1+tan2x-tan2x))[sec4x+sec2xtan2x+(sec2x-1)tan2x]

 

                   = sec4x - tan2x + 2sec2xtan2x

 

                   = sec2x(1+tan2x) - tan2x + 2sec2xtan2x

 

                   = sec2x - tan2x + sec2xtan2x + 2sec2xtan2x

 

                   = 1 + 3sec2xtan2x

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hope this helps you dear