Cho biết $\cot \alpha = 5$. Tính giá trị của $E = 2{\cos ^2}\alpha + 5\sin \alpha \cos \alpha  + 1$?

Ta có:

$\begin{array}{l} E = 2{\cos ^2}\alpha + 5\sin \alpha \cos \alpha + 1\\ = {\sin ^2}\alpha \left( {2.\frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} + 5.\frac{{\cos \alpha }}{{\sin \alpha }} + \frac{1}{{{{\sin }^2}\alpha }}} \right)\\ = \left( {\frac{1}{{{{\cot }^2}\alpha + 1}}} \right)\left( {2{{\cot }^2}\alpha + 5\cot \alpha + \frac{1}{{\frac{1}{{{{\cot }^2}\alpha + 1}}}}} \right)\\ = \left( {\frac{1}{{{{\cot }^2}\alpha + 1}}} \right)\left( {2{{\cot }^2}\alpha + 5\cot \alpha + {{\cot }^2}\alpha + 1} \right)\\ = \left( {\frac{1}{{{{\cot }^2}\alpha + 1}}} \right)\left( {3{{\cot }^2}\alpha + 5\cot \alpha + 1} \right)\\ = \frac{1}{{{5^2} + 1}}\left( {{{3.5}^2} + 5.5 + 1} \right) = \frac{{101}}{{26}} \end{array}$