Question

chứng minh: \(\frac{tanA}{tanB}=\frac{c^2+a^2-b^2}{c^2+b^2-a^2}\)

Answer 1

Ta có: \(a^2=b^2+c^2-2bc\cos A\)

\(\Leftrightarrow\cos A=\frac{b^2+c^2-a^2}{2bc}\)

Tương tự: \(\Leftrightarrow\cos B=\frac{a^2+c^2-b^2}{2ac}\)

Ta lại có: \(\frac{a}{\sin A}=\frac{b}{\sin B}=2R\)

\(\left\{\begin{matrix}\sin A=\frac{a}{2R}\\\sin B=\frac{b}{2R}\end{matrix}\right.\)

Quay lại bài toán ta có:

\(\frac{\tan A}{\tan B}=\frac{\sin A\cos B}{\sin B\cos A}=\frac{\frac{a}{2R}.\frac{a^2+c^2-b^2}{2ac}}{\frac{b}{2R}.\frac{b^2+c^2-a^2}{2bc}}=\frac{c^2+a^2-b^2}{c^2+b^2-a^2}\)