Nếu \(\tan \left( \alpha  \right)\) và \(\tan \left( \beta  \right)\) là nghiệm của phương trình \({x^2} - px + q = 0,{\rm{ }}(q \ne 1)\) thì giá trị của biểu thức \(Q = {\cos ^2}\left( {\alpha  + \beta } \right) + p\sin \left( {\alpha  + \beta } \right)\cos \left( {\alpha  + \beta } \right) + q{\sin ^2}\left( {\alpha  + \beta } \right)\) bằng

Ta có \(\tan \left( \alpha  \right)\) và \(\tan \left( \beta  \right)\) là nghiệm của phương trình \({x^2} - px + q = 0,{\rm{ }}(q \ne 1)\)

Theo định lí Vi-ét ta có:

\(\left\{ \begin{array}{l}\tan \left( \alpha  \right) + \tan \left( \beta  \right) = p\\\tan \left( \alpha  \right).\tan \left( \beta  \right) = q\end{array} \right. \Rightarrow \tan \left( {\alpha  + \beta } \right) = \frac{{\tan \left( \alpha  \right) + \tan \left( \beta  \right)}}{{1 - \tan \left( \alpha  \right).\tan \left( \beta  \right)}} = \frac{p}{{1 - q}}\)\(\begin{array}{l} \Rightarrow {\cos ^2}\left( {\alpha  + \beta } \right) = \frac{1}{{1 + {{\tan }^2}\left( {\alpha  + \beta } \right)}} = \frac{1}{{1 + \frac{{{p^2}}}{{{{(1 - q)}^2}}}}} = \frac{{{{(1 - q)}^2}}}{{{{(1 - q)}^2} + {p^2}}}\\q \ne 1 \Rightarrow \frac{{\sin \left( \alpha  \right)\sin \left( \beta  \right)}}{{\cos \left( \alpha  \right)\cos \left( \beta  \right)}} \ne 1 \Rightarrow \sin \left( \alpha  \right)\sin \left( \beta  \right) \ne \cos \left( \alpha  \right)\cos \left( \beta  \right)\\ \Rightarrow \cos \left( {\alpha  + \beta } \right) = \cos \left( \alpha  \right)\cos \left( \beta  \right) - \sin \left( \alpha  \right)\sin \left( \beta  \right) \ne 0\\ \Rightarrow Q = {\cos ^2}\left( {\alpha  + \beta } \right)\left[ {1 + p.\frac{{\sin \left( {\alpha  + \beta } \right)}}{{\cos \left( {\alpha  + \beta } \right)}} + q.\frac{{{{\sin }^2}\left( {\alpha  + \beta } \right)}}{{{{\cos }^2}\left( {\alpha  + \beta } \right)}}} \right]\\ = {\cos ^2}\left( {\alpha  + \beta } \right)\left[ {1 + p.\tan \left( {\alpha  + \beta } \right) + q.{{\tan }^2}\left( {\alpha  + \beta } \right)} \right]\\ = \frac{{{{(1 - q)}^2}}}{{{{(1 - q)}^2} + {p^2}}}\left[ {1 + \frac{{{p^2}}}{{1 - q}} + \frac{{{p^2}q}}{{{{\left( {1 - q} \right)}^2}}}} \right] = \frac{{{{(1 - q)}^2}}}{{{{(1 - q)}^2} + {p^2}}}.\frac{{{{(1 - q)}^2} + {p^2}(1 - q) + {p^2}q}}{{{{(1 - q)}^2}}}\\ = 1\end{array}\)