Đáp án:
B
Giải thích các bước giải:
\(\begin{array}{l}
\tan \alpha = \sqrt 2 \\
\to \frac{{\sin \alpha }}{{\cos \alpha }} = \sqrt 2 \\
\to \sin \alpha = \sqrt 2 \cos \alpha \\
Có:{\sin ^2}\alpha + {\cos ^2}\alpha = 1\\
\to {\left( {\sqrt 2 \cos \alpha } \right)^2} + {\cos ^2}\alpha = 1\\
\to 2{\cos ^2}\alpha + {\cos ^2}\alpha = 1\\
\to {\cos ^2}\alpha = \frac{1}{3}\\
C = \frac{{3 - {{\cos }^2}\alpha }}{{5{{\cos }^2}\alpha - 2{{\left( {\sqrt 2 \cos \alpha } \right)}^2} + 1}}\\
= \frac{{3 - {{\cos }^2}\alpha }}{{5{{\cos }^2}\alpha - 2\left( {2.{{\cos }^2}\alpha } \right) + 1}}\\
= \frac{{3 - {{\cos }^2}\alpha }}{{5{{\cos }^2}\alpha - 4{{\cos }^2}\alpha + 1}}\\
= \frac{{3 - {{\cos }^2}\alpha }}{{{{\cos }^2}\alpha + 1}} = \frac{{3 - \frac{1}{3}}}{{\frac{1}{3} + 1}} = 2
\end{array}\)
