Answer
$\cot(x+y)=\frac{\cot x\cot y-1}{\cot x+\cot y}$
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 addition 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 $\cot x\cot y$:
$=\frac{(1-\frac{1}{\cot x}\frac{1}{\cot y})\cot x\cot y}{(\frac{1}{\cot x}+\frac{1}{\cot y})\cot x\cot y}$
$=\frac{\cot x\cot y-1}{\cot y+\cot x}$
$=\frac{\cot x\cot y-1}{\cot x+\cot y}$
Since this equals the right side, the identity has been proven.
