Question

In a triangle $ABC$, prove that
$\frac{\tan A}{\tan B}=\frac{c^2+a^2-b^2}{c^2+b^2-a^2}$

Answer 1

$\frac{\tan A}{\tan B}=\frac{c^2+a^2-b^2}{c^2+b^2-a^2}$

$\frac{c^2+a^2-b^2}{2ac}=\cos B$

$\frac{c^2+b^2-a^2}{2bc}=\cos A$

$\frac{\frac{\frac{c^2+a^2-b^2}{2abc}}{c^2+b^2-a^2}}{2abc}=\frac{\frac{\frac{\cos B}{b}}{\cos A}}{a}$

$=\frac{\cos B}{b}\times \frac{a}{\cos A}$

$=\frac{2R\sin A}{2R\sin B}\frac{\cos B}{\cos A}$

$=\frac{\tan A}{\tan B}$

Hence proved$.$