Question

[Thầy Tùng Dương]

Cho $\tan \alpha = 3$. Tính

a) \(\dfrac{2\sin\alpha+3\cos\alpha}{3\sin\alpha-4\cos\alpha}.\)

b) \(\dfrac{\sin\alpha\cos\alpha}{\sin^2\alpha-\sin\alpha\cos\alpha+\cos^2\alpha}.\)

Answer 1

[NGUYỄN VĂN TUẤN VIỆT]

a) \(\dfrac{2sina+3cosa}{3sina-4cosa}=\dfrac{9}{5}\)

b) \(\dfrac{sina.cosa}{sin^2a-sina.cosa+cos^2a}=0\)

Answer 2

[NGUYỄN TÀI ANH]

​\(a.\dfrac{2\sin\alpha+3\cos\alpha}{3\sin\alpha-4\cos\alpha}=\dfrac{2\left(3cos\alpha\right)+3cos\alpha}{3\left(3cos\alpha\right)-4cos\alpha}=\dfrac{9cos\alpha}{5cos\alpha}=\dfrac{9}{5}\)
\(b.\dfrac{sin\alpha cos\alpha}{sin^2\alpha-sin\alpha cos\alpha+cos^2\alpha}=\dfrac{3cos^2\alpha}{9cos^2\alpha-3cos^2\alpha+cos^2\alpha}=\dfrac{3cos^2\alpha}{7cos^2\alpha}=\dfrac{3}{7}\)