Xét dạng tam giác ABC thỏa mãn : $\frac{1+cosB}{SinB}$ = $\frac{2a+c}{\sqrt{4a^2-c^2}}$

Đáp án:

$\Delta ABC$ cân tại $C.$

Giải thích các bước giải:

$\dfrac{1+\cos B}{\sin B}=\dfrac{2a+c}{\sqrt{4a^2-c^2}}\\ \Rightarrow \left(\dfrac{1+\cos B}{\sin B}\right)^2=\left(\dfrac{2a+c}{\sqrt{4a^2-c^2}}\right)^2\\ \Leftrightarrow \dfrac{(1+\cos B)^2}{\sin^2B}=\dfrac{(2a+c)^2}{4a^2-c^2}\\ \Leftrightarrow \dfrac{(1+\cos B)^2}{1-\cos^2B}=\dfrac{(2a+c)^2}{(2a+c)(2a-c)}\\ \Leftrightarrow \dfrac{(1+\cos B)^2}{(1+\cos B)(1-\cos B)}=\dfrac{2a+c}{2a-c}\\ \Leftrightarrow \dfrac{1+\cos B}{1-\cos B}=\dfrac{2a+c}{2a-c}\\ \Leftrightarrow \dfrac{1+\cos B}{1-\cos B}=\dfrac{2 .2R\sin A+2R\sin C}{2.2R\sin A-2R\sin C}\\ \Leftrightarrow \dfrac{1+\cos B}{1-\cos B}=\dfrac{2\sin A+\sin C}{2\sin A-\sin C}\\ \Rightarrow (1+\cos B)(2\sin A-\sin C)= (1-\cos B)(2\sin A+\sin C)\\ \Leftrightarrow 2\sin A-\sin C+2\sin A\cos B-\sin C\cos B=2\sin A+\sin C -2\sin A\cos B-\sin C\cos B\\ \Leftrightarrow -\sin C+2\sin A\cos B=\sin C -2\sin A\cos B\\ \Leftrightarrow 2\sin C -4\sin A\cos B=0\\ \Leftrightarrow \sin C -2\sin A\cos B=0\\ \Leftrightarrow \sin C -(\sin( A+B)+\sin(A- B))=0\\ \Leftrightarrow \sin C -\sin( 180^\circ-C)-\sin(A- B)=0\\ \Leftrightarrow \sin C -\sin(C)-\sin(A- B)=0\\ \Leftrightarrow -\sin(A- B)=0\\ \Leftrightarrow A- B=0\\ \Leftrightarrow A= B$

$\Rightarrow \Delta ABC$ cân tại $C.$