Cho biết $\tan \alpha =-\frac{3}{4},{90}^{°}<\alpha <{180}^{°}$. Khi đó:

a) \(\cos \alpha  > 0\)

b) \(\cos \alpha  =  - \frac{4}{5}\)

c) \(\cot \alpha  =  - \frac{4}{3}\)

d) \(\sin \alpha  =  - \frac{3}{5}\)

\({\cos ^2}\alpha  = 1 - {\sin ^2}\alpha  = \frac{{144}}{{169}} \Rightarrow \cos \alpha  =  \pm \frac{{12}}{{13}}\)

Do \(\alpha \) là góc tù nên \(\cos \alpha  < 0\), từ đó \(\cos \alpha  =  - \frac{{12}}{{13}}\)

Như vậy \(3\sin \alpha  + 2\cos \alpha  = 3 \cdot \frac{5}{{13}} + 2\left( { - \frac{{12}}{{13}}} \right) =  - \frac{9}{{13}}\)

Vì \(\cot \alpha  = 2 \Rightarrow \sin \alpha  \ne 0\). Chia cả tử và mẫu cho \({\sin ^3}\alpha \) ta được:

\(B = \frac{{(\sin \alpha  + 2\cos \alpha )\frac{1}{{{{\sin }^3}\alpha }}}}{{\left( {{{\sin }^3}\alpha  - {{\cos }^3}\alpha } \right)\frac{1}{{{{\sin }^3}\alpha }}}} = \frac{{\frac{1}{{{{\sin }^2}\alpha }} + 2\cot \alpha  \cdot \frac{1}{{{{\sin }^2}\alpha }}}}{{1 - {{\cot }^3}\alpha }}\)\(\)

\( = \frac{{1 + {{\cot }^2}\alpha  + 2\cot \alpha \left( {1 + {{\cot }^2}\alpha } \right)}}{{1 - {{\cot }^3}\alpha }} = \frac{{2{{\cot }^3}\alpha  + {{\cot }^2}\alpha  + 2\cot \alpha  + 1}}{{1 - {{\cot }^3}\alpha }} =  - \frac{{25}}{7}{\rm{. }}\)