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) \(\sin \alpha  =  - \frac{3}{5}\)

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

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

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

a) \(A = 4 \cdot \frac{1}{2} + 3\sqrt 3  \cdot \frac{1}{{\sqrt 3 }} = 2 + 3 = 5\).

b) \(B = \frac{1}{{\sqrt 3 }} \cdot \frac{{\sqrt 3 }}{2} - 3\sqrt 2  \cdot \frac{{\sqrt 2 }}{2} + 1 = \frac{1}{2} - 3 + 1 =  - \frac{3}{2}\).

c) \(C = {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} + {\left( {\frac{1}{{\sqrt 3 }}} \right)^2} - 2 = \frac{3}{4} + \frac{1}{3} - 2 =  - \frac{{11}}{{12}}\).

d) $D=2\sin {30}^{°}+2\sqrt{3}-\frac{1}{2\sqrt{2}}\cos {45}^{°}+3\tan {30}^{°}$ do $\left\{\begin{array}{l}\begin{array}{l}\sin {150}^{°}=\sin \left({180}^{°}-{30}^{°}\right)=\sin {30}^{°}\\ \cos {135}^{°}=\cos \left({180}^{°}-{45}^{°}\right)=-\cos {45}^{°}\\ \tan {150}^{°}=\tan \left({180}^{°}-{30}^{°}\right)=-\tan {30}^{°}\end{array}\end{array}\right.$

$D=2\cdot \frac{1}{2}+2\sqrt{3}-\frac{1}{2\sqrt{2}}\cdot \frac{\sqrt{2}}{2}+3\cdot \frac{\sqrt{3}}{3}=1+2\sqrt{3}-\frac{1}{4}+\sqrt{3}=\frac{3}{4}+3\sqrt{3}.$