If y=(sinx-cosx)/(sinx+cosx)y=sinxcosxsinx+cosx, then find the value of 1+y^21+y2?

1 Answer
May 13, 2017

1+y^2=2/(1+sin2x)1+y2=21+sin2x

Explanation:

As y=(sinx-cosx)/(sinx+cosx)y=sinxcosxsinx+cosx

1+y^2=1+((sinx-cosx)/(sinx+cosx))^21+y2=1+(sinxcosxsinx+cosx)2

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

=(2sin^2x+2cos^2x)/(sin^2x+cos^2x+2sinxcosx)=2sin2x+2cos2xsin2x+cos2x+2sinxcosx

=2/(1+sin2x)=21+sin2x