What you wrote is this:


\sin x \ = \ \cos x \ = \frac {2 sin^2 x} {9 \sin x - \cos x}


Is that really what you meant?

I meant this (sinx+cosx)=(2sin^2x-1)/(sinx-cosx)?

I meant (sinx+cosx)=(2sin^2x)/(sinx-cosx)

I ment this (sinx+cosx)=(2sin^2x-1)/(sinx-cosx)?

OK:


\sin x + \cos x = \frac { \sin^2x-1}{\sin x - \cos x}


Here's a hint to get you started: multiply through by the denominator of the right hand side. This turns the left hand side into


(\sin x + \cos x )(\sin x - \cos x) = \sin^2 x - cos^2 x


Now think about the various identities for cos (2x). OK, that's enough of a hint - can you take it from here?