Answer
$1-\tan x\tan y=\frac{\cos(x+y)}{\cos x\cos y}$
Work Step by Step
Start with the right side:
$\frac{\cos(x+y)}{\cos x\cos y}$
Expand using the addition formula for sine:
$=\frac{\cos x\cos y-\sin x\sin y}{\cos x\cos y}$
Simplify:
$=\frac{\cos x\cos y}{\cos x\cos y}-\frac{\sin x\sin y}{\cos x\cos y}$
$=1-\frac{\sin x}{\cos x}\frac{\sin y}{\cos y}$
$=1-\tan x\tan y$
Since this equals the left side, the identity has been proven.
