Xét tính chất của tam giác $ABC$ biết rằng: $\cos A + \cos B - \cos C + 1 = \sin A + \sin B + \sin C$

Ta có:

$\cos A + \cos B - \cos C + 1$ $ = 2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2} + 2{\sin ^2}\dfrac{C}{2}$$ = 2\cos \left( {\dfrac{\pi }{2} - \dfrac{C}{2}} \right)\cos \dfrac{{A - B}}{2} + 2{\sin ^2}\dfrac{C}{2}$ $ = 2\sin \dfrac{C}{2}\cos \dfrac{{A - B}}{2} + 2{\sin ^2}\dfrac{C}{2}$$ = 2\sin \dfrac{C}{2}\left( {\cos \dfrac{{A - B}}{2} + \sin \dfrac{C}{2}} \right)$ $ = 2\sin \dfrac{C}{2}\left( {\cos \dfrac{{A - B}}{2} + \cos \dfrac{{A + B}}{2}} \right)$ $ = 2\sin \dfrac{C}{2}.2\cos \dfrac{A}{2}.\cos \dfrac{B}{2}$ $ = 4\cos \dfrac{A}{2}\cos \dfrac{B}{2}\sin \dfrac{C}{2}$

$\sin A + \sin B + \sin C$ $ = 2\sin \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2} + 2\sin \dfrac{C}{2}\cos \dfrac{C}{2}$ $ = 2\sin \left( {\dfrac{\pi }{2} - \dfrac{C}{2}} \right)\cos \dfrac{{A - B}}{2} + 2\sin \dfrac{C}{2}\cos \dfrac{C}{2}$ $ = 2\cos \dfrac{C}{2}\cos \dfrac{{A - B}}{2} + 2\sin \dfrac{C}{2}\cos \dfrac{C}{2}$ $ = 2\cos \dfrac{C}{2}\left( {\cos \dfrac{{A - B}}{2} + \sin \left( {\dfrac{\pi }{2} - \dfrac{{A + B}}{2}} \right)} \right)$ $ = 2\cos \dfrac{C}{2}\left( {\cos \dfrac{{A - B}}{2} + \cos \dfrac{{A + B}}{2}} \right)$

$ = 2\cos \dfrac{C}{2}.2\cos \dfrac{A}{2}\cos \dfrac{B}{2}$ $ = 4\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2}$

$\dfrac{{\cos A + \cos B - \cos C + 1}}{{\sin A + \sin B + \sin C}}$ $ = \dfrac{{4\cos \dfrac{A}{2}\cos \dfrac{B}{2}\sin \dfrac{C}{2}}}{{4\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2}}} = \tan \dfrac{C}{2}$

$ \Rightarrow \tan \dfrac{C}{2} = 1 \Leftrightarrow \dfrac{C}{2} = {45^0} \Leftrightarrow C = {90^0}$

$ \Rightarrow \Delta ABC$ vuông tại $C.$