Answer
See the explanation below.
Work Step by Step
One of the sum-to-product formulas is $\sin \alpha -\sin \beta =2\sin \frac{\alpha -\beta }{2}\cos \frac{\alpha +\beta }{2}$. Therefore, $\sin x-\sin y$ can be written as shown below:
$\sin x-\sin y=2\sin \frac{x-y}{2}\cos \frac{x+y}{2}$
One of the sum-to-product formulas is $\cos \alpha -\cos \beta =-2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$. Therefore, $\cos x-\cos y$ can be written as shown below:
$\cos x-\cos y=-2\sin \frac{x+y}{2}\sin \frac{x-y}{2}$
Now, consider the left side of the provided expression:
$\frac{\sin x-\sin y}{\cos x-\cos y}$
The expression can be simplified as provide below:
$\begin{align}
& \frac{\sin x-\sin y}{\cos x-\cos y}=\frac{2\sin \left( \frac{x-y}{2} \right)\cos \left( \frac{x+y}{2} \right)}{-2\sin \left( \frac{x+y}{2} \right)\sin \left( \frac{x-y}{2} \right)} \\
& =-\frac{\sin \left( \frac{x-y}{2} \right)\cos \left( \frac{x+y}{2} \right)}{\sin \left( \frac{x+y}{2} \right)\sin \left( \frac{x-y}{2} \right)} \\
& =-\frac{\sin \left( \frac{x-y}{2} \right)}{\sin \left( \frac{x-y}{2} \right)}\cdot \frac{\cos \left( \frac{x+y}{2} \right)}{\sin \left( \frac{x+y}{2} \right)} \\
& =-\frac{\cos \left( \frac{x+y}{2} \right)}{\sin \left( \frac{x+y}{2} \right)}
\end{align}$
Then, by using one of the quotient identities of trigonometry, which is $\cot x=\frac{\cos x}{\sin x}$, the expression can be further written as:
$-\frac{\cos \left( \frac{x+y}{2} \right)}{\sin \left( \frac{x+y}{2} \right)}=-\cot \frac{x+y}{2}$
Thus, the left side of the expression is equal to the right side, which is
$\frac{\sin x-\sin y}{\cos x-\cos y}=-\cot \frac{x+y}{2}$.
