Cho sin α = 12/ 13 ( 0 ∘ < α < 90 ∘ ) . Khi đó: a) cos α < 0

Vì 0°<α<90° nên \(\cos \alpha  > 0\).

Do đó \(\cos \alpha  = \sqrt {1 - {{\sin }^2}\alpha }  = \frac{5}{{13}};\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{12}}{5};\cot \alpha  = \frac{5}{{12}}\).