Cho góc \(\alpha\) thỏa mãn \(\sin\alpha=\dfrac{3}{5}\) và \(\dfrac{\pi}{2}<\alpha<\pi\). Tính \(P=\dfrac{\tan\alpha}{1+\tan^2\alpha}\).
![]() | \(P=-3\) |
![]() | \(P=\dfrac{3}{7}\) |
![]() | \(P=\dfrac{12}{25}\) |
![]() | \(P=-\dfrac{12}{25}\) |
Chọn phương án D.
Vì \(\dfrac{\pi}{2}<\alpha<\pi\) nên \(\cos\alpha<0\).
Ta có \(\cos^2\alpha=1-\sin^2\alpha=1-\dfrac{9}{25}=\dfrac{16}{25}\).
Suy ra \(\cos\alpha=-\dfrac{4}{5}\).
Khi đó \(\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{\dfrac{3}{5}}{-\dfrac{4}{5}}=-\dfrac{3}{4}\).
Vậy \(P=\dfrac{\tan\alpha}{1+\tan^2\alpha}=\dfrac{-\dfrac{3}{4}}{1+\dfrac{9}{16}}=-\dfrac{12}{25}\).