Question

Verify sec^2 theta + tan^2 theta = (1-sin^4theta)sec^4theta

Answer 1

\[\bf \sec^2(\theta)+\tan^2(\theta)=(1-\sin^4(\theta))\sec^4(\theta)\]Re-write the right side by expanding first: \[\bf (1-\sin^4(\theta))\sec^4(\theta)=\sec^4(\theta)-\sin^4(\theta)\sec^4(\theta)=\sec^4(\theta)-\tan^4(\theta)\]\[\bf = [\sec^2(\theta)+\tan^2(\theta)][\sec^2(\theta)-\tan^2(\theta)]\]\[\bf = [\sec^2(\theta)+\tan^2(\theta)]\left[ \frac{ 1 }{ \cos^2(\theta) } -\frac{ \sin^2(\theta) }{ \cos^2(\theta) }\right]=[\sec^2(\theta)+\tan^2(\theta)]\left( \frac{ \cos^2(\theta) }{ \cos^2(\theta) } \right)\]\[\bf =\sec^2(\theta)+\tan^2(\theta) \implies R.S=L.S\]

Answer 2

Do you see how that works?

@longhornsjason

Answer 3

Awesome thank you!

Answer 4

@longhornsjason You do understand right?

Answer 5

Yes, this makes sense. thank you. I was stuck on this for awhile.