Answer
See the full explanation below.
Work Step by Step
$\frac{\cot x+\cot y}{1-\cot x\cot y}$
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
$\frac{\cot x+\cot y}{1-\cot x\cot y}=\frac{\cos x\sin y+\sin x\cos y}{\sin x\sin y-\cos x\cos y}$.
