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Question

$nếu a^{2}+b^{2}=c^{2}+4R^{2} thì \frac{(\tan A\tan B+1)}{(\tan A\tan B-1)}=(\tan C)^{2}$

khangnguyenthanh · Answer 1 · 10/15/2012 6:46:18 PM

Ta cÓ:

$a^2+b^2=c^2+4R^2$

$\Leftrightarrow 4R^2\sin^2A+4R^2\sin^2B=4R^2\sin^2C+4R^2$

$\Leftrightarrow \sin^2A+\sin^2B=\sin^2C+1$

$\Leftrightarrow 2\sin^2A-1+2\sin^2B-1=2\sin^2C$

$\Leftrightarrow -\cos2A-\cos2B=2\sin^2C$

$\Leftrightarrow -\cos(A-B)\cos(A+B)=\sin^2C$

$\Rightarrow \tan^2C=\frac{-\cos(A-B)\cos(A+B)}{\cos^2C}$

$=\frac{-\cos(A-B)}{\cos(A+B)}$

$=\frac{\sin A\sin B+\cos A\cos B}{\sin A\sin B-\cos A\cos B}$

$=\frac{\tan A\tan B+1}{\tan A\tan B-1}$

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