Prove the following: (sec^(6)A+tan^(6)A)/(sec^(2)A+tan^(2)A) = 1+sec^(2)A*tan^(2)A
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Step-by-step explanation:
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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Answer:
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