Ta có:
\(1+tan^2x=\dfrac{1}{cos^2x}\)
\(\Leftrightarrow cos^2x=\dfrac{1}{1+tan^2x}\)
\(\Leftrightarrow cos^2x=\dfrac{1}{1+3^2}\)
\(\Leftrightarrow cosx=\sqrt{\dfrac{1}{10}}=\dfrac{\sqrt{10}}{10}\)
Mà: \(tanx=\dfrac{sinx}{cosx}\)
\(\Leftrightarrow sinx=tanx\cdot cosx\)
\(\Leftrightarrow sinx=3\cdot\dfrac{\sqrt{10}}{10}=\dfrac{3\sqrt{10}}{10}\)
Giá trị của A là:
\(A=\dfrac{\dfrac{3\sqrt{10}}{10}+4\cdot\dfrac{\sqrt{10}}{10}}{2\cdot\dfrac{3\sqrt{10}}{10}-\dfrac{\sqrt{10}}{10}}\)
\(A=\dfrac{\dfrac{3\sqrt{10}}{10}+\dfrac{4\sqrt{10}}{10}}{\dfrac{6\sqrt{10}}{10}-\dfrac{\sqrt{10}}{10}}\)
\(A=\dfrac{\dfrac{7\sqrt{10}}{10}}{\dfrac{5\sqrt{10}}{10}}\)
\(A=\dfrac{7}{5}\)
tan=3
=>sin=3*cos
\(A=\dfrac{sin+4cos}{2sin-cos}=\dfrac{3cos+4cos}{6cos-cos}=\dfrac{7}{5}\)
