Step 1: Understand the data.
We know $\theta + \phi = \alpha$ and $\tan\theta = k\tan\phi$ with $k>1$. We want $\sin(\theta-\phi)$ in terms of $\alpha$ and $k$.
Step 2: Turn the tangent condition into sines and cosines.
Write $\tan\theta = k\tan\phi$ as \[ \frac{\sin\theta}{\cos\theta} = k\,\frac{\sin\phi}{\cos\phi}, \] which gives $\sin\theta\cos\phi = k\sin\phi\cos\theta$.
Step 3: Use componendo and dividendo.
From $\frac{\sin\theta\cos\phi}{\sin\phi\cos\theta} = \frac{k}{1}$, apply the rule that turns $\frac{a}{b}=\frac{k}{1}$ into $\frac{a+b}{a-b}=\frac{k+1}{k-1}$. So \[ \frac{\sin\theta\cos\phi + \cos\theta\sin\phi}{\sin\theta\cos\phi - \cos\theta\sin\phi} = \frac{k+1}{k-1}. \]
Step 4: Recognise the sum and difference formulas.
The numerator is $\sin(\theta+\phi)$ and the denominator is $\sin(\theta-\phi)$. So \[ \frac{\sin(\theta+\phi)}{\sin(\theta-\phi)} = \frac{k+1}{k-1}. \]
Step 5: Put in $\theta+\phi=\alpha$.
Since $\theta+\phi=\alpha$, \[ \frac{\sin\alpha}{\sin(\theta-\phi)} = \frac{k+1}{k-1}. \]
Step 6: Solve for $\sin(\theta-\phi)$.
Cross multiply and isolate: \[ \sin(\theta-\phi) = \left(\frac{k-1}{k+1}\right)\sin\alpha. \] \[ \boxed{\sin(\theta-\phi) = \left(\dfrac{k-1}{k+1}\right)\sin\alpha} \]