Triangle question, proving isosceles given trigonometric conditions

Question

$ABC$ is a triangle satisfying the following condition: $$\frac{\sin B}{\sin A}=\frac{\tan B+\cot C}{\tan A+\cot C}$$

How do I prove that $ABC$ is isosceles? I really have no idea.

Answer 1

$$\require{cancel} \frac{\sin B}{\sin A}=\frac{\tan B+\cot C}{\tan A+\cot C}=\frac{\sin B\sin C+\cos C\cos B}{\sin A\sin C+\cos C\cos A}.\frac{\cos A}{\cos B}=\frac{\cos(B-C)}{\cos(A-C)}.\frac{\cos A}{\cos B}\\ \implies\sin 2A\cos(B-C)=\sin 2B\cos(A-C)\\ \implies\sin(2A+B-C)+\sin(2A-B+C)=\sin(2B+A-C)+\sin(2B-A+C)\\ \implies\sin(\pi+A-C)+\sin(\pi+A-B)=\sin(\pi+B-C)+\sin(\pi+B-A)\\ \implies\sin(A-C)+\sin(A-B)=\sin(B-C)+\sin(B-A)\\ \implies\sin(A-C)-\sin(B-C)=2\sin(B-A)\\ \implies\cos\frac{A+B-2C}2\sin\frac{A-B}2=2\sin\frac{B-A}2\cos\frac{B-A}2\\ \implies\cos\frac{\pi-3C}2+2\cos\frac{A-B}2=0\\ \cdots \text{continue for other solutions} $$ We can see $\sin\frac{A-B}2$ terms on both sides there must exist other solutions if $\sin\frac{A-B}2\ne0$.Otherwise anyways $\sin\frac{A-B}2$ is a solution or $A=B$ is a solution.