$\begin{cases}sinα+cosα=\dfrac{7}{5}\\sin^2α+cos^2α=1\\\end{cases}$
`<=>` $\begin{cases}sinα+cosα=\dfrac{7}{5}\\(sinα+cosα)^2-2sinαcosα=1\\\end{cases}$
`<=>` $\begin{cases}sinα+cosα=\dfrac{7}{5}\\sinα.cosα=\dfrac{12}{25}\\\end{cases}$
`<=>` \(\left\{{}\begin{matrix}\left[{}\begin{matrix}sinα=\dfrac{4}{5}\\cosα=\dfrac{3}{5}\end{matrix}\right.\\\left[{}\begin{matrix}sinα=\dfrac{3}{5}\\cosα=\dfrac{4}{5}\end{matrix}\right.\end{matrix}\right.\)
`=>` \(\left[{}\begin{matrix}tanα=\dfrac{3}{4}\\tanα=\dfrac{4}{3}\end{matrix}\right.\)
Vậy...
Ta có: \(\left(\sin\alpha+\cos\alpha\right)^2=\dfrac{49}{25}\)
\(\Leftrightarrow2\cdot\sin\alpha\cdot\cos\alpha=\dfrac{49}{25}-1=\dfrac{24}{25}\)
Ta có: \(\left(\sin\alpha-\cos\alpha\right)^2\)
\(=\sin^2\alpha+\cos^2\alpha-\dfrac{24}{25}\)
\(=1-\dfrac{24}{25}=\dfrac{1}{25}\)
\(\Leftrightarrow\sin\alpha-\cos\alpha=\dfrac{1}{5}\)
mà \(\sin\alpha+\cos\alpha=\dfrac{7}{5}\)
nên \(2\cdot\sin\alpha=\dfrac{8}{5}\)
hay \(\sin\alpha=\dfrac{4}{5}\)
\(\Leftrightarrow\cos\alpha=\dfrac{7}{5}-\dfrac{4}{5}=\dfrac{3}{5}\)
\(\Leftrightarrow\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\)