Question

hwy plz answer dis
prove that

cosx by 1-tan x - sin square x by cosx - sinx

Answer 1

$\frac{cosx}{1-tanx}-\frac{sin^2x}{cosx-sinx}\\\\=\frac{cosx}{1-\frac{sinx}{cosx}}-\frac{sin^2x}{cosx-sinx}\\\\=\frac{cos^2x}{cosx-sinx}-\frac{sin^2x}{cosx-sinx}\\\\=\frac{cos^2x-sin^2x}{cosx-sinx}\\\\=\frac{(cosx-sinx)(cosx+sinx)}{cosx-sinx}\\\\=cosx+sinx,\ \ \ if\ cosx \ne sinx$