Cho biết $\tan \alpha =-\frac{3}{4},{90}^{°}<\alpha <{180}^{°}$. Khi đó:

a) \(\cos \alpha  > 0\)

b) \(\cos \alpha  =  - \frac{4}{5}\)

c) \(\cot \alpha  =  - \frac{4}{3}\)

d) \(\sin \alpha  =  - \frac{3}{5}\)

\({\cos ^2}\alpha  = 1 - {\sin ^2}\alpha  = \frac{{144}}{{169}} \Rightarrow \cos \alpha  =  \pm \frac{{12}}{{13}}\)

Do \(\alpha \) là góc tù nên \(\cos \alpha  < 0\), từ đó \(\cos \alpha  =  - \frac{{12}}{{13}}\)

Như vậy \(3\sin \alpha  + 2\cos \alpha  = 3 \cdot \frac{5}{{13}} + 2\left( { - \frac{{12}}{{13}}} \right) =  - \frac{9}{{13}}\)

Chọn C

Ta có \(1 + {\tan ^2}\alpha  = \frac{1}{{{{\cos }^2}\alpha }} \Leftrightarrow {\cos ^2}\alpha  = \frac{1}{{1 + {{\tan }^2}\alpha }} = \frac{1}{{1 + {3^2}}} = \frac{1}{{10}}\).

Suy ra \(\cos \alpha  =  \pm \frac{{\sqrt {10} }}{{10}}\).