How do you prove (1 - sin2x) /(cos2x) = (cos2x) / (1 + sin2x)?

1 Answer
May 14, 2015

We have to prove that (1-sin2x)/(cos2x)=(cos2x)/(1+sin2x)

To do this we transform left side:

(1-sin2x)/(cos2x)=((sin^2x+cos^2x)-2sinxcosx)/(cos2x)

=(sin^2x-2sinxcosx+cos^2x)/(cos^2x-sin^2x)

=(sinx-cosx)^2/((cosx-sinx)(cosx+sinx))

=((sinx-cosx)^2)/(-(sinx-cosx)(sinx+cosx))

=(cosx-sinx)/(cosx+sinx)

Now we expand the expresion by multiplying both numerator and denominator by (cosx+sinx)

So we get:

((cosx-sinx)(cosx+sinx))/((cosx+sinx)^2)

=(cos^2x-sin^2x)/(cos^2x+2cosxsinx+sin^2x)

=(cos2x)/(1+sin2x)