## Precalculus (6th Edition) Blitzer

The solution of the given question is $\cos x-1$.
$\frac{\cos x\tan x-\tan x+2\cos x-2}{\tan x+2}$ By rearranging the numerator so that the numerator can further be separated, we get: $\frac{\cos x\tan x-\tan x+2\cos x-2}{\tan x+2}=\frac{\cos x\tan x+2\cos x-\tan x-2}{\tan x+2}$ Now, the denominator can be written separately for the numerator in the following expression: \begin{align} & \frac{\cos x\tan x+2\cos x-\tan x-2}{\tan x+2}=\frac{\cos x\tan x+2\cos x}{\tan x+2}+\frac{-\tan x-2}{\tan x+2} \\ & =\frac{\cos x\left( \tan x+2 \right)}{\tan x+2}-1 \\ & =\cos x-1 \end{align} Conjecture: Left side is equal to $\cos x-1$. Thus, the left side of the expression is equal to the right side, which is $\frac{\cos x\tan x-\tan x+2cox-2}{\tan x+2}=\cos x-1$.