If an angle $\alpha$ is divided into two parts A and B such that $A-B=x$ and $\tan A : \tan B=k:1$ then $\sin x=$

$$\begin{gathered} A-B=x,A+B=\alpha . \\ \frac{\tan A}{\tan B}=\frac{k}{1}\Rightarrow \frac{\sin A\cos B}{\cos A\sin B}=\frac{k}{1}u\sin g componendo and dividendo \\ \Rightarrow \frac{\sin A\cos B+\cos A\sin B}{\sin A\cos B-\cos A\sin B}=\frac{k+1}{k-1} \\ \Rightarrow \frac{\sin \left(A+B\right)}{\sin (A-B)}=\frac{k+1}{k-1} \\ \Rightarrow \frac{\sin \alpha }{\sin x}=\frac{k+1}{k-1} \\ \Rightarrow \sin x=\frac{k-1}{k+1}\sin \alpha \end{gathered}$$