## Precalculus (6th Edition) Blitzer

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 given below: $\sin x-\sin y=2\sin \frac{x-y}{2}\cos \frac{x+y}{2}$ 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 given below: $\sin x+\sin y=2\sin \frac{x+y}{2}\cos \frac{x-y}{2}$ Now, consider the left side of the given expression: $\frac{\sin x-\sin y}{\sin x+\sin y}$ The expression can be simplified as shown below: \begin{align} & \frac{\sin x-\sin y}{\sin x+\sin 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)\cos \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)\cos \left( \frac{x-y}{2} \right)} \\ & =\frac{\sin \left( \frac{x-y}{2} \right)}{\cos \left( \frac{x-y}{2} \right)}\cdot \frac{\cos \left( \frac{x+y}{2} \right)}{\sin \left( \frac{x+y}{2} \right)} \end{align} The quotient identities of trigonometry are $\tan x=\frac{\sin x}{\cos x}$ and $\cot x=\frac{\cos x}{\sin x}$. Therefore, by applying these identities, the expression can be further simplified as: $\frac{\sin \left( \frac{x-y}{2} \right)}{\cos \left( \frac{x-y}{2} \right)}\cdot \frac{\cos \left( \frac{x+y}{2} \right)}{\sin \left( \frac{x+y}{2} \right)}=\tan \frac{x-y}{2}\cot \frac{x+y}{2}$ Thus, the left side of the expression is equal to the right side, which is $\frac{\sin x-\sin y}{\sin x+\sin y}=\tan \frac{x-y}{2}\cot \frac{x+y}{2}$