Answer
See the full explanation below.
Work Step by Step
$\frac{\tan x+\tan y}{1-\tan x\tan y}$
Apply the quotient identity of trigonometry $\tan x=\frac{\sin x}{\cos x}$ and $\tan y=\frac{\sin y}{\cos y}$. Then the above expression can be further simplified as:
$\frac{\tan x+\tan y}{1-\tan x\tan y}=\frac{\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}}{1-\frac{\sin x}{\cos x}.\frac{\sin y}{\cos y}}$
By expressing the terms in the numerator and denominator with least common denominator $\cos x\cos y$, we get:
$\frac{\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}}{1-\frac{\sin x}{\cos x}.\frac{\sin y}{\cos y}}=\frac{\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y}}{\frac{\cos x\cos y-\sin x\sin y}{\cos x\cos y}}$
Then rewrite the main fraction bar as $\div $:
$\frac{\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y}}{\frac{\cos x\cos y-\sin x\sin y}{\cos x\cos y}}=\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y}\div \frac{\cos x\cos y-\sin x\sin y}{\cos x\cos y}$
Invert the divisor and multiply:
$\begin{align}
& \frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y}\div \frac{\cos x\cos y-\sin x\sin y}{\cos x\cos y}=\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y}.\frac{\cos x\cos y}{\cos x\cos y-\sin x\sin y} \\
& =\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}
\end{align}$
Thus, the left side of the expression is equal to the right side, which is
$\frac{\tan x+\tan y}{1-\tan x\tan y}=\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}$.
