Question

Biết \(tan\frac{\alpha}{2}=2\). tính \(A=\frac{1+5cos\alpha}{3-2cos\alpha}\)

Answer 1

\(1+tan^2\frac{a}{2}=\frac{1}{cos^2\frac{a}{2}}\Rightarrow cos^2\frac{a}{2}=\frac{1}{1+tan^2\frac{a}{2}}=\frac{1}{5}\)

\(A=\frac{1+5\left(2cos^2\frac{a}{2}-1\right)}{3-2\left(2cos^2\frac{a}{2}-1\right)}=\frac{-4+10cos^2\frac{a}{2}}{5-4cos^2\frac{a}{2}}=\frac{-4+10.\frac{1}{5}}{5-4.\frac{1}{5}}=...\)