Cho cos α = 2 3 . Tính A = tan α + 3 cot α/tan α + cot α .

\(A = \frac{{\tan \alpha + 3\cot \alpha }}{{\tan \alpha + \cot \alpha }}\)

\( = \frac{{\tan \alpha + 3 \cdot \frac{1}{{\tan \alpha }}}}{{\tan \alpha + \frac{1}{{\tan \alpha }}}}\)

\( = \frac{{{{\tan }^2}\alpha + 3}}{{{{\tan }^2}\alpha + 1}}\)

\( = \frac{{\frac{1}{{{{\cos }^2}x}} + 2}}{{\frac{1}{{{{\cos }^2}x}}}}\)

\( = 1 + 2{\cos ^2}x\).

Mà \(\cos \alpha = \frac{2}{3}\) nên \(A = 1 + 2 \cdot {\left( {\frac{2}{3}} \right)^2} = \frac{{17}}{9}\).