Ta cÓ:
$a^2+b^2=c^2+4R^2$
$\Leftrightarrow 4R^2\sin^2A+4R^2\sin^2B=4R^2\sin^2C+4R^2$
$\Leftrightarrow \sin^2A+\sin^2B=\sin^2C+1$
$\Leftrightarrow 2\sin^2A-1+2\sin^2B-1=2\sin^2C$
$\Leftrightarrow -\cos2A-\cos2B=2\sin^2C$
$\Leftrightarrow -\cos(A-B)\cos(A+B)=\sin^2C$
$\Rightarrow \tan^2C=\frac{-\cos(A-B)\cos(A+B)}{\cos^2C}$
$=\frac{-\cos(A-B)}{\cos(A+B)}$
$=\frac{\sin A\sin B+\cos A\cos B}{\sin A\sin B-\cos A\cos B}$
$=\frac{\tan A\tan B+1}{\tan A\tan B-1}$