Question

Trigonometry
sec2x=? check all that apply
A. 1/cos^2x-sin^2x
B.1/1-2sin^2x
C.1/sin^2x-cos^2x
D.1/2sin^2x-1

Answer 1

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

Answer 2

Thanks

Answer 3

NP