Question

Cho \(0^o< \alpha< 90^o\) và \(\sin\alpha+\cos\alpha=\frac{7}{5}\). Tính \(\tan\alpha\).

Answer 1

\(\sin^2\alpha+\cos^2\alpha=1\)

\(\Rightarrow\sin^2\alpha+\left(\frac{7}{5}-\sin\alpha\right)^2=1\)

\(\Rightarrow25\sin^2\alpha-35\sin\alpha+12=0\)

\(\Rightarrow\left(5\sin\alpha-4\right)\left(5\sin\alpha-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\sin\alpha=\frac{4}{5}\\\sin\alpha=\frac{3}{5}\end{cases}}\)

Nếu \(\sin\alpha=\frac{4}{5}\)thì \(\cos\alpha=\frac{3}{5}\Rightarrow\tan\alpha=\frac{4}{3}\)

Nếu \(\sin\alpha=\frac{3}{5}\)thì \(\cos\alpha=\frac{4}{5}\Rightarrow\tan\alpha=\frac{3}{4}\)

Tk cho mk bạn nhá