Question

sec 2x=
check all that apply
1/cos^2x-sin^2x
1/1-2sin^2x
1/sin^2x-cos^2x
1/2sin^2x-1
I know that sec=1/cos

Answer 1

\[\sec 2x=\frac{ 1 }{ \cos 2x }=\frac{ 1 }{ \cos ^2x-\sin ^2x }=\frac{ 1 }{ 1-2\sin ^2x }\]

Answer 2

Got it, since 1-sin^2x=cos^2x so it is the same as 1/1-sin^2x=sec2x which is 1/cos^2x
but 1/cos^2x-sin^2x doesnt make sense to how it would work here?

Answer 3

\[\cos 2x=\cos \left( x+x \right)=\cos x \cos x-\sin x \sin x=\cos ^2x-\sin ^2x\]