1. Cho tam giác ABC
a) CMR nếu cos (A+C)+3cosB =1 thì góc B= 60do
b) CMR nếu [tex]\frac{1+cosB}{sinB}[/tex]=[tex]\frac{2a+c}{\sqrt{4a^{2}-c^{2}}}[/tex] thì tam giác ABC cân.
a.
[tex]cos (A+C)+3.cosB =1\\\Rightarrow -cosB+3.cosB=1\\\Rightarrow cosB=\frac{1}{2}\\\Rightarrow B=60[/tex]
b.
[tex]\frac{1 + \cos B}{\sin B} = \frac{2a + c}{\sqrt{4a^{2} - c^2}}\\\Rightarrow \frac{1 + \cos B}{\sin B} = \frac{2a + c}{\sqrt{(2a + c)(2a - c)}}\\ \Rightarrow \frac{1 + \cos B}{\sin B} = \sqrt{\frac{2a + c}{2a - c}} \\\Rightarrow \frac{(1+cosB)^2}{sin^2B}=\frac{2a + c}{2a - c}\\\Rightarrow \frac{(1+cosB)^2}{1-cos^2B}=\frac{2a + c}{2a - c}\\\Rightarrow \frac{1+cosB}{1-cosB}=\frac{2a + c}{2a - c}\\\Rightarrow cosB=\frac{c}{2a}\\\Rightarrow \frac{a^2+c^2-b^2}{2ac}=\frac{c}{2a}\\\Rightarrow a^2-b^2=0\\\Rightarrow (a-b)(a+b)=0[/tex]
Nên a=b (Do a+b luôn >0)