## Precalculus (6th Edition) Blitzer

$\frac{\cot x+\cot y}{1-\cot x\cot y}$ Using the quotient identity of trigonometry $\cot x=\frac{\cos x}{\sin x}$ and $\cot y=\frac{\cos y}{\sin y}$, then the above expression can be further simplified as: $\frac{\cot x+\cot y}{1-\cot x\cot y}=\frac{\frac{\cos x}{\sin x}+\frac{\cos y}{\sin y}}{1-\frac{\cos x}{\sin x}.\frac{\cos y}{\sin y}}$ By expressing the terms in the numerator and denominator with least common denominator $\sin x\sin y$, we get: $\frac{\frac{\cos x}{\sin x}+\frac{\cos y}{\sin y}}{1-\frac{\cos x}{\sin x}.\frac{\cos y}{\sin y}}=\frac{\frac{\cos x\sin y+\sin x\cos y}{\sin x\sin y}}{\frac{\sin x\sin y-\cos x\cos y}{\sin x\sin y}}$ Then rewrite the main fraction bar as $\div$: $\frac{\frac{\cos x\sin y+\sin x\cos y}{\sin x\sin y}}{\frac{\sin x\sin y-\cos x\cos y}{\sin x\sin y}}=\frac{\cos x\sin y+\sin x\cos y}{\sin x\sin y}\div \frac{\sin x\sin y-\cos x\cos y}{\sin x\sin y}$ After that, invert the divisor and multiply: \begin{align} & \frac{\cos x\sin y+\sin x\cos y}{\sin x\sin y}\div \frac{\sin x\sin y-\cos x\cos y}{\sin x\sin y}=\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y}.\frac{\cos x\cos y}{\cos x\cos y-\sin x\sin y} \\ & =\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y} \end{align} Thus, the left side of the expression is equal to the right side, which is $\frac{\cot x+\cot y}{1-\cot x\cot y}=\frac{\cos x\sin y+\sin x\cos y}{\sin x\sin y-\cos x\cos y}$.