`(sin ( theta -phi))/(sin theta sin phi) = cot phi - cot theta.`

Answer

To prove the equation

\[ \frac{\sin(\theta - \phi)}{\sin \theta \sin \phi} = \cot \phi - \cot \theta, \]

we will follow these steps:

### Step 1: Use the sine difference identity
We start with the left-hand side, which is \(\frac{\sin(\theta - \phi)}{\sin \theta \sin \phi}\). We will use the sine difference identity:

\[ \sin(\theta - \phi) = \sin \theta \cos \phi - \cos \theta \sin \phi. \]

Substituting this into our expression gives:

\[ \frac{\sin \theta \cos \phi - \cos \theta \sin \phi}{\sin \theta \sin \phi}. \]

### Step 2: Separate the fraction
Now, we can separate the fraction into two parts:

\[ \frac{\sin \theta \cos \phi}{\sin \theta \sin \phi} - \frac{\cos \theta \sin \phi}{\sin \theta \sin \phi}. \]

This simplifies to:

\[ \frac{\cos \phi}{\sin \phi} - \frac{\cos \theta}{\sin \theta}. \]

### Step 3: Recognize cotangent
We know that:

\[ \frac{\cos x}{\sin x} = \cot x. \]

Thus, we can rewrite our expression as:

\[ \cot \phi - \cot \theta. \]

### Conclusion
Now we have shown that:

\[ \frac{\sin(\theta - \phi)}{\sin \theta \sin \phi} = \cot \phi - \cot \theta. \]

Hence, the equation is proved.

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