Answer
$\dfrac{\tan x+\tan y}{\cot x+\cot y}=\tan x\tan y$
Work Step by Step
$\dfrac{\tan x+\tan y}{\cot x+\cot y}=\tan x\tan y$
Replace $\tan x$ and $\tan y$ by $\dfrac{\sin x}{\cos x}$ and $\dfrac{\sin y}{\cos y}$, respectively. Also replace $\cot x$ and $\cot y$ by $\dfrac{\cos x}{\sin x}$ and $\dfrac{\cos y}{\sin y}$, respectively.
$\dfrac{\dfrac{\sin x}{\cos x}+\dfrac{\sin y}{\cos y}}{\dfrac{\cos x}{\sin x}+\dfrac{\cos y}{\sin y}}=\tan x\tan y$
Evaluate the sums indicated on the left side:
$\dfrac{\dfrac{\sin x\cos y+\sin y\cos x}{\cos x\cos y}}{\dfrac{\sin y\cos x+\sin x\cos y}{\sin x\sin y}}=\tan x\tan y$
Evaluate the division and simplify:
$\dfrac{(\sin x\cos y+\sin y\cos x)(\sin x\sin y)}{(\sin y\cos x+\sin x\cos y)(\cos x\cos y)}=\tan x\tan y$
$\dfrac{\sin x\sin y}{\cos x\cos y}=\tan x\tan y$
Since $\dfrac{\sin x}{\cos x}=\tan x$ and $\dfrac{\sin y}{\cos y}=\tan y$, the identity is proved.
$\tan x\tan y=\tan x\tan y$
