1)Ta có:\(\sin2\alpha=2\sin\alpha.\cos\alpha\)
\(1+\cos2\alpha=2\cos^2\alpha\)
\(\Rightarrow A=\frac{2\sin\alpha.\cos\alpha+\sin\alpha}{2\cos^2\alpha+\cos\alpha}=\frac{\sin\alpha\left(2\cos\alpha+1\right)}{\cos\alpha\left(2\cos\alpha+1\right)}=\frac{\sin\alpha}{\cos\alpha}=\tan\alpha\)
\(B=\sin\left(\frac{\pi}{3}+\alpha\right)-\sin\left(\frac{\pi}{3}-\alpha\right)\)
\(B=\frac{\sqrt{3}}{2}.\cos\alpha+\frac{1}{2}\sin\alpha-\frac{\sqrt{3}}{2}.\cos\alpha+\frac{1}{2}\sin\alpha=\sin\alpha\)
(Sử dụng công thức biến đổi \(\sin\left(\alpha\pm\beta\right)=\sin\alpha.\cos\beta\pm\cos\alpha.\sin\beta\))
