Ta có : \(S=\dfrac{1}{2}SinB.ac=b^2-a^2-c^2+2ac\)
\(\Rightarrow\dfrac{1}{2}SinB.ac=-\left(a^2+c^2-b^2\right)+2ac\)
Mà \(CosB=\dfrac{a^2+c^2-b^2}{2ac}\)
\(\Rightarrow a^2+c^2-b^2=2ac.CosB\)
\(\Rightarrow\dfrac{1}{2}SinB.ac=2ac-2ac.\cos B\)
\(\Rightarrow SinB=4-4\cos B\)
\(\Rightarrow SinB+4\cos B=4\)
Lại có : \(\sin^2B+\cos^2B=1\)
- Giair hệ ta được : \(\left\{{}\begin{matrix}\left[{}\begin{matrix}\cos B=1\\\cos B=\dfrac{15}{17}\end{matrix}\right.\\\left[{}\begin{matrix}sinB=0\\sinB=\dfrac{8}{17}\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}tanB=0\\tanB=\dfrac{8}{15}\end{matrix}\right.\)
Mà 3 điểm A, B, C là 1 tam giác .
=> TanB = 8/15 .
