Cho cos α = − 2/5 ( 90 ∘ < α < 180 ∘ ) . Tính tan α .
Giải thích
\(1 + {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }} \Rightarrow {\tan ^2}\alpha = \frac{1}{{{{\cos }^2}\alpha }} - 1 = \frac{1}{{\frac{4}{{25}}}} - 1 = \frac{{21}}{4}\).
\( \Rightarrow \tan \alpha = - \frac{{\sqrt {21} }}{2}\) (Do 90°<α<180°