prove that sin^2x-tan^2x=-sin^2xtan^2x

Question

I need the answer asap

Answer 1

sin^2x-tan^2x=-sin^2xtan^2x

working with the left side to make it look like the right side...

sin^2x-(sin^2x/cos^2x)

(sin^2xcos^2x)/cos^2x-(sin^2x/cos^2x)

combine the fractions with a common denominator

(sin^2xcos^2x-sin^2x)cos^2x

factor the numerator

sin^2x(cos^2x-1)/cos^2x

sin^2x+cos^2x=1, so...

cos^2x-1=-sin^2x

(sin^2x)(-sin^2x)/cos^2x

(-sin^2x)(sin^2x/cos^2x)

(-sin^2x)(tan^2x) which is the right side so we proved the identity (don't forget that tan=sin/cos so tan^2=sin^2/cos^2)