Cho góc \(\alpha\) thỏa mãn \(\cos\alpha=-\dfrac{12}{13}\) và \(\dfrac{\pi}{2}<\alpha<\pi\). Tính \(\tan\alpha\).
\(\tan\alpha=-\dfrac{12}{5}\) | |
\(\tan\alpha=\dfrac{5}{12}\) | |
\(\tan\alpha=-\dfrac{5}{12}\) | |
\(\tan\alpha=\dfrac{12}{5}\) |
Chọn phương án C.
Vì \(\dfrac{\pi}{2}<\alpha<\pi\) nên \(\tan\alpha<0\).
Ta có \(1+\tan^2\alpha=\dfrac{1}{\cos^2\alpha}=\dfrac{169}{144}\).
Suy ra \(\tan^2\alpha=\dfrac{169}{144}-1=\dfrac{25}{144}\).
Vậy \(\tan\alpha=-\dfrac{5}{12}\).