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View Full Version : Sec^6x-tan^6x = 1 3(tan^2x)(sec^2x)


Lilp22
Dec 5, 2010, 08:11 AM
We're proving identifies in class, and I was kind confused on how to prove this one.

Lilp22
Dec 5, 2010, 09:17 AM
Sorry there was suppose to be a plus in that question
sec^6x-tan^6x = 1 + 3(tan^2x)(sec^2x)

neerajtiwari
Feb 18, 2012, 10:53 PM
Left Hand Side = sec^6 x - tan^6 x
= (sec²x - tan²x)(sec^4 x + (tan²x)(sec²x) + tan^4 x)
= 1*(sec^4 x + (tan²x)(sec²x) + tan^4 x)
= sec^4 x + (tan²x)(sec²x) + (tan²x)²

= sec^4 x + (tan²x)(sec²x) + (sec²x - 1)²
= sec^4 x + (tan²x)(sec²x) + sec^4 x - 2sec²x + 1
= 1 + 2sec^4 x - 2sec²x + (tan²x)(sec²x)

= 1 + 2sec²x(sec²x - 1) + (tan²x)(sec²x)
= 1 + 2(sec²x)(tan²x) + (tan²x)(sec²x)
= 1 + 3(tan²x)(sec²x) = Right Hand Side