Ta có P=$\frac{sin^4a-cos^4a-sin^2a}{sin^2a+sina}$= $\frac{(sin^2a+cos^2a)(sin^2a-cos^2a)-sin^2a}{sina(1+sina)}$=$\frac{sin^2a-cos^2a-sin^2a}{sina(1+sina)}$= $\frac{-cos^2a}{sina(1+sina)}$=$\frac{-(1-sin^2a)}{sina(1+sina)}$ =$\frac{-(1+sina)(1-sina)}{sina(1+sina)}$= $\frac{-(1-sina)}{sina}$= $\frac{sina-1}{sina}$= $\frac{sina}{sina}$- $\frac{1}{sina}$= 1+$\frac{-1}{sina}$
Theo bài ra ta có: P=a+$\frac{b}{sina}$ ⇒ a=1, b=-1 ⇒ a+b=0 ⇒A