Answer
$\cot(x-y)=\frac{\cot x\cot y+1}{\cot y-\cot x}$
Work Step by Step
Start with the left side:
$\cot(x-y)$
Express cotangent as 1 divided by tangent:
$=\frac{1}{\tan(x-y)}$
Expand using the subtraction formula for tangent:
$=\frac{1}{\frac{\tan x-\tan y}{1+\tan x\tan y}}$
Multiply the top and bottom by ${1+\tan x\tan y}$:
$=\frac{1*(1+\tan x\tan y)}{\frac{\tan x-\tan y}{1+\tan x\tan y}*(1+\tan x\tan y)}$
$=\frac{1+\tan x\tan y}{\tan x-\tan y}$
Express tangent as 1 divided by cotangent:
$=\frac{1+\frac{1}{\cot x}\frac{1}{\cot y}}{\frac{1}{\cot x}-\frac{1}{\cot y}}$
Multiply top and bottom by $\cot x\cot y$:
$=\frac{(1+\frac{1}{\cot x}\frac{1}{\cot y})\cot x\cot y}{(\frac{1}{\cot x}-\frac{1}{\cot y})\cot x\cot y}$
$=\frac{\cot x\cot y+1}{\cot y-\cot x}$
Since this equals the right side, the identity has been proven.
