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

1 Answer
Daniel L. · mason m
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)#

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