a: \(A=\dfrac{\left(sina+cosa\right)\left(sina-cosa\right)}{\left(sina+cosa\right)^2}=\dfrac{sina-cosa}{sina+cosa}\)
b: tan a=2/3 nên sin a/cosa=2/3
=>sina=2/3cosa
\(A=\dfrac{\dfrac{2}{3}cosa-cosa}{\dfrac{2}{3}cosa+cosa}=\dfrac{-1}{3}:\dfrac{5}{3}=\dfrac{-1}{5}\)