Cho biết tan α = − 3 4 , 90 ∘ < α < 180 ∘ . Khi đó: a) cos α > 0

Vì \(\tan \alpha  =  - \frac{3}{4} \Rightarrow \cot \alpha  = \frac{1}{{\tan \alpha }} =  - \frac{4}{3}\).

Theo giả thiết: 90°<α<180°⇒cosα<0

Ta có: \(\frac{1}{{{{\cos }^2}\alpha }} = 1 + {\tan ^2}\alpha  = 1 + {\left( { - \frac{3}{4}} \right)^2} = \frac{{25}}{{16}} \Rightarrow {\cos ^2}\alpha  = \frac{{16}}{{25}} \Rightarrow \cos \alpha  =  - \frac{4}{5}\); \(\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} \Rightarrow \sin \alpha  = \tan \alpha  \cdot \cos \alpha  =  - \frac{3}{4} \cdot \left( { - \frac{4}{5}} \right) = \frac{3}{5}\).