Cho biết \(\cos \alpha = - \frac{2}{3}\). Tính \(\tan \alpha \)? 

Vì \(\cot \alpha  =  - 3\)nên \(\sin \alpha  \ne 0.\) Chia cả tử và mẫu của \(P\) cho \({\sin ^3}\alpha \), ta có:

\(P = \frac{{1 + {{\cot }^3}\alpha }}{{\frac{1}{{{{\sin }^2}\alpha }}\left( {1 - \cot \alpha } \right)}} = \frac{{1 + {{\cot }^3}\alpha }}{{\left( {1 + {{\cot }^2}\alpha } \right)\left( {1 - \cot \alpha } \right)}} = \frac{{1 + {{\left( { - 3} \right)}^3}}}{{\left[ {1 + {{\left( { - 3} \right)}^2}} \right]\left[ {1 - \left( { - 3} \right)} \right]}} =  - \frac{{13}}{{20}}\)

a) \(D = a.1 + b.0 + c.0 = a\).

b) \(E = {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} + 2 \cdot {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} - 5.1 =  - \frac{{11}}{4}\).

c) \(F = {\cos ^2}{24^0} + {\sin ^2}{24^0} + {\cos ^2}{1^0} + {\sin ^2}{1^0} = 1 + 1 = 2\).

d) $G={\cos }^2{45}^{°}+{\sin }^2{45}^{°}+{\sin }^2{45}^{°}-2\left({\cos }^2{50}^{°}+{\cos }^2{40}^{°}\right)-5\tan {55}^{°}\cot {55}^{°}$

$={\cos }^2{45}^{°}+{\sin }^2{45}^{°}+{\sin }^2{45}^{°}-2\left({\cos }^2{50}^{°}+{\sin }^2{50}^{°}\right)-5\tan {55}^{°}\cot {55}^{°}=\frac{-11}{2}$