Đáp án:
$\dfrac{{11}}{{13}}$
Giải thích các bước giải:
Ta có:
$\tan B = 2 \Rightarrow \left\{ \begin{array}{l}
\cos B \ne 0\\
\dfrac{1}{{{{\cos }^2}B}} = {\tan ^2}B + 1
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{\cos ^2}B \ne 0\\
\dfrac{1}{{{{\cos }^2}B}} = 5
\end{array} \right.$
Khi đó:
$\begin{array}{l}
\dfrac{{{{\sin }^2}B + 2{{\cos }^2}B + 1}}{{{{\sin }^2}B - {{\cos }^2}B + 2}}\\
= \dfrac{{\dfrac{{{{\sin }^2}B}}{{{{\cos }^2}B}} + 2 + \dfrac{1}{{{{\cos }^2}B}}}}{{\dfrac{{{{\sin }^2}B}}{{{{\cos }^2}B}} - 1 + \dfrac{2}{{{{\cos }^2}B}}}}\\
= \dfrac{{{{\tan }^2}B + 2 + 5}}{{{{\tan }^2}B - 1 + 2.5}}\\
= \dfrac{{4 + 2 + 5}}{{4 - 1 + 10}}\\
= \dfrac{{11}}{{13}}
\end{array}$
