Question

\(A=\frac{sin\alpha-cos\alpha}{sin\alpha+cos\alpha}\)

tính A khi \(tan\alpha\)=\(\frac{1}{3}\)

Answer 1

Bạn thang Tran hình như bị nhầm?

\(\frac{sina-cosa}{sina+cosa}=\frac{cosa.tana-cosa}{cosa.tana+cosa}=\frac{cosa\left(tana-1\right)}{cosa\left(tana+1\right)}=\frac{tana-1}{tana+1}\)

Answer 2

A = \(\frac{sina-cosa}{sina+cosa}=\frac{1-tana}{1+tana}=\frac{1-\frac{1}{3}}{1+\frac{1}{3}}=\frac{\frac{2}{3}}{\frac{4}{3}}=\frac{2}{3}\cdot\frac{3}{4}=\frac{1}{2}\)