Xét tính chất của tam giác ABC biết rằng: cosA + cosB – cosC + 1 = sinA + sinB

* Hướng dẫn giải

Đáp án C

Ta có:

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

$$\begin{gathered} \sin A+\sin B+\sin C=2\sin \frac{A+B}{2}\cos \frac{A-B}{2}+2\sin \frac{C}{2}\cos \frac{C}{2} \\ =2\sin \left(\frac{\pi }{2}-\frac{C}{2}\right)\cos \frac{A-B}{2}+2\sin \frac{C}{2}\cos \frac{C}{2} \\ =2\cos \frac{C}{2}\cos \frac{A-B}{2}+2\sin \frac{C}{2}\cos \frac{C}{2} \\ =2\cos \frac{C}{2}\left(\cos \frac{A-B}{2}+\sin \left(\frac{\pi }{2}-\frac{A+B}{2}\right)\right) \\ =2\cos \frac{C}{2}\left(\cos \frac{A-B}{2}+\cos \frac{A+B}{2}\right) \\ =2\cos \frac{C}{2}.2\cos \frac{A}{2}\cos \frac{B}{2} \\ =4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} \\ \Rightarrow \frac{\cos A+\cos B+\cos C+1}{\sin A+\sin B+\sin C} \\ =\frac{4\cos \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2}}{4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}}=\tan \frac{C}{2} \\ \Rightarrow \tan \frac{C}{2}=1\iff \frac{C}{2}={45}^0\iff C={90}^0 \end{gathered}$$

$\Rightarrow \Delta ABC$ vuông tại C