Phần II. Trắc nghiệm đúng, sai

Cho \(\sin \alpha = \frac{3}{5},\left( {90^\circ < \alpha < 180^\circ } \right)\).

a) \(\cos \alpha  > 0\).

b) \({\cos ^2}\alpha  = \frac{{16}}{{25}}\).

c) \[\tan \left( {180^\circ  - \alpha } \right) =  - \frac{3}{4}\].

d) \[A = \frac{{\tan \alpha  - \cot \left( {180^\circ  - \alpha } \right)}}{{\sin \left( {90^\circ  - \alpha } \right)}} = \frac{{125}}{{48}}\].

a) Sai. Ta có \(90^\circ  < \alpha  < 180^\circ \) nên \(\cos \alpha  < 0\).

b) Đúng. Vì \({\sin ^2}\alpha  + {\cos ^2}\alpha  = 1 \Rightarrow {\cos ^2}\alpha  = 1 - {\sin ^2}\alpha  = 1 - {\left( {\frac{3}{5}} \right)^2} = \frac{{16}}{{25}}\).

Do đó \[\cos \alpha  =  - \sqrt {\frac{{16}}{{25}}}  =  - \frac{4}{5}\].

c) Sai. Ta có \[\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} =  - \frac{3}{4} \Rightarrow \,\tan \left( {180^\circ  - \alpha } \right) =  - \tan \alpha  = \frac{3}{4}\].

d) Đúng. \[A = \frac{{\tan \alpha  - \cot \left( {180^\circ  - \alpha } \right)}}{{\sin \left( {90^\circ  - \alpha } \right)}} = \frac{{\tan \alpha  - \frac{1}{{\tan \left( {180^\circ  - \alpha } \right)}}}}{{\cos \alpha }} = \frac{{\frac{{ - 3}}{4} - \frac{4}{3}}}{{\frac{{ - 4}}{5}}} = \frac{{125}}{{48}}\].