If \(\theta+\phi=\alpha\) and \(\tan \theta=\mathrm{k} \tan \phi\) (where \(\mathrm{K}>1)\), then the value of \(\sin (\theta-\phi)\) is

We have \(\tan \theta=\mathrm{k} \tan \phi\) and \(\theta+\phi=\alpha\)
\(
\therefore \frac{\tan \theta}{\tan \phi}=\frac{k}{1}
\)
By Componendo Dividendo, we get
\(
\begin{aligned}
& \frac{\tan \theta+\tan \phi}{\tan \theta-\tan \phi}=\frac{\mathrm{k}+1}{\mathrm{k}-1} \\
& \therefore \frac{\frac{\sin \theta}{\cos \theta}+\frac{\sin \phi}{\cos \phi}}{\frac{\sin \theta}{\cos \theta}-\frac{\sin \phi}{\cos \phi}}=\frac{k+1}{k-1} \\
& \therefore \frac{\sin \cos \phi+\cos \theta \sin \phi}{\sin \theta \cos \phi-\cos \theta \sin \phi}=\frac{k+1}{k-1} \\
& \therefore \frac{\sin (\theta+\phi)}{\sin (\theta-\phi)}=\frac{\mathrm{k}+1}{\mathrm{k}-1} \Rightarrow \frac{\sin \alpha}{\sin (\theta-\phi)}=\frac{\mathrm{k}+1}{\mathrm{k}-1} \\
& \therefore \sin (\theta-\phi)=\frac{\mathrm{k}-1}{\mathrm{k}+1}(\sin \alpha) \\
&
\end{aligned}
\)