Answer
$\dfrac{\cos(x+y)}{\cos x\sin y}=\cot y-\tan x$
Work Step by Step
$\dfrac{\cos(x+y)}{\cos x\sin y}=\cot y-\tan x$
Substitute $\cos(x+y)$ by $\cos x\cos y-\sin x\sin y$:
(This is the addition formula for cosine)
$\dfrac{\cos x\cos y-\sin x\sin y}{\cos x\sin y}=\cot y-\tan x$
Rewrite the left side as $\dfrac{\cos x\cos y}{\cos x\sin y}-\dfrac{\sin x\sin y}{\cos x\sin y}$ and simplify:
$\dfrac{\cos x\cos y}{\cos x\sin y}-\dfrac{\sin x\sin y}{\cos x\sin y}=\cot y-\tan x$
$\dfrac{\cos y}{\sin y}-\dfrac{\sin x}{\cos x}=\cot y-\tan x$
Since $\dfrac{\cos y}{\sin y}=\cot y$ and $\dfrac{\sin x}{\cos x}=\tan x$, the identity is proved.
$\cot y-\tan x=\cot y-\tan x$
