Cho $A, B, C$ là $3$ góc của một tam giác.a) Chứng minh:  $\frac{\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}   }{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}}=1.$b) Giả sử $\cos C(\sin A+\sin B)=\sin C.\cos(A-B).$  Hãy tính $\cos A+\cos B$.

a) Trong mọi tam giác $ABC$ đều có:
$\tan (\frac{A+B}{2})=\cot \frac{C}{2}=\frac{1}{\tan \frac{C}{2} }$, suy ra $\frac{\tan \frac{A}{2}+\tan \frac{B}{2}  }{1-\tan \frac{A}{2}\tan \frac{B}{2} }=\frac{1}{\tan \frac{C}{2} }$
Từ đó: $\tan \frac{A}{2}\tan \frac{B}{2}+\tan \frac{B}{2}\tan \frac{C}{2}+\tan \frac{C}{2}\tan \frac{A}{2}=1$
$\Leftrightarrow \frac{1}{\cot \frac{A}{2}\cot \frac{B}{2}}+\frac{1}{\cot \frac{B}{2}\cot \frac{C}{2} }+\frac{1}{\cot \frac{C}{2}\cot \frac{A}{2} }=1$
$\Leftrightarrow \frac{\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}   }{\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}}=1.$

b) Giả thiết $\cos C(\sin A+\sin B)=\sin C.\cos(A-B)$ tương đương với:
$2 \cos C \sin \frac{A+B}{2} \cos \frac{A-B}{2}=2 \sin \frac{C}{2}\cos \frac{C}{2}(2 \cos^2 \frac{A-B}{2}-1)$
$\Leftrightarrow \cos C \cos \frac{A-B}{2}=\sin \frac{C}{2}(2 \cos^2 \frac{A-B}{2}-1)$
$\Leftrightarrow (1-2 \sin^2 \frac{C}{2}) \cos \frac{A-B}{2}=\sin \frac{C}{2}(2 \cos^2 \frac{A-B}{2}-1)$
$\Leftrightarrow \cos \frac{A-B}{2}+\sin \frac{C}{2}=2 \sin^2 \frac{C}{2} \cos \frac{A-B}{2}+2 \sin \frac{C}{2}\cos^2 \frac{A-B}{2}$
$=2 \sin \frac{C}{2} \cos \frac{A-B}{2}( \cos \frac{A-B}{2}+\sin \frac{C}{2}).$
Vì $\cos \frac{A-B}{2}+\sin \frac{C}{2}\neq  0$ nên đẳng thức trên tương đương với:
$2 \sin \frac{C}{2} \cos \frac{A-B}{2}=1\Leftrightarrow 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}=1$
$\Leftrightarrow \cos A+\cos B=1$