Verified Answer :

Ryan2

Remember:

\(\boxed{sin(2x)=2.sinx.cosx}\\ \\ and\\ \\ \boxed{cos(2x)=cos^2x-sin^2x}\)


\(\frac{2cos2x}{sin2x}-cotx-tanx=\\ \\ \frac{2(cos^2x-sin^2x)}{2sinx.cosx}-\frac{cosx}{sinx}-\frac{sinx}{cosx}=\\ \\ \frac{cos^2x-sin^2x}{sinx.cosx}-\frac{cosx}{sinx}-\frac{sinx}{cosx}=\\ \\ \frac{cos^2x-sin^2x-cos^2x-sin^2x}{sinx.cosx}=\\ \\ \frac{-2sin^2x}{sinx.cosx}=\\ \\ \frac{-2sinx}{cosx}=-2.tanx\)