Cho biết \(\sin \alpha = \frac{3}{5},\frac{\pi }{2} < \alpha < \pi \). Khi đó:

a) \(\cos \alpha < 0\)

b) \(\cos \alpha = - \frac{4}{5}\)

c) \(\tan \alpha = \frac{3}{4}\)

d) \(\tan \left( {\alpha + \frac{\pi }{3}} \right) = \frac{{48 - \sqrt 3 }}{{11}}\)

\(\begin{array}{l}{\rm{ V\`i  }}\frac{\pi }{2} < \alpha  < \pi  \Rightarrow \cos \alpha  < 0.{\rm{ }}\\{\rm{Ta c\'o  }}\cos \alpha  =  - \sqrt {1 - {{\sin }^2}\alpha }  =  - \frac{4}{5}\\ \Rightarrow \tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }} =  - \frac{3}{4}{\rm{. }}\end{array}\)

Ta có: \(\tan \left( {\alpha  + \frac{\pi }{3}} \right) = \frac{{\tan \alpha  + \tan \frac{\pi }{3}}}{{1 - \tan \alpha \tan \frac{\pi }{3}}} = \frac{{\tan \alpha  + \sqrt 3 }}{{1 - \sqrt 3 \tan \alpha }}\).

Suy ra:\(\tan \left( {\alpha  + \frac{\pi }{3}} \right) = \frac{{ - \frac{3}{4} + \sqrt 3 }}{{1 - \sqrt 3 \left( { - \frac{3}{4}} \right)}} = \frac{{ - 3 + 4\sqrt 3 }}{{4 + 3\sqrt 3 }} = \frac{{48 - 25\sqrt 3 }}{{11}}\).