Problem
\( \frac{\tan 2 x-\tan x}{1+\tan 2 x \tan x}= \)
Solution
Step 1 ：\( \frac{\tan 2 x-\tan x}{1+\tan 2 x \tan x}= \frac{\frac{2 \tan x}{1-\tan^2 x} - \tan x}{1+\frac{2 \tan x}{1-\tan^2 x} \tan x} \)
Step 2 ：\( \frac{\frac{2 \tan x}{1-\tan^2 x} - \tan x}{1+\frac{2 \tan x}{1-\tan^2 x} \tan x} = \frac{2 \tan x - \tan x (1-\tan^2 x)}{1+ 2 \tan^2 x - \tan^4 x} \)
Step 3 ：\( \frac{2 \tan x - \tan x (1-\tan^2 x)}{1+ 2 \tan^2 x - \tan^4 x} = \frac{\tan x (2-1+\tan^2 x)}{1-\tan^2 x + 2 \tan^2 x} \)
Step 4 ：\( \frac{\tan x (2-1+\tan^2 x)}{1-\tan^2 x + 2 \tan^2 x} = \frac{\tan x (1+\tan^2 x)}{1+ \tan^2 x} \)
Step 5 ：\( \frac{\tan x (1+\tan^2 x)}{1+ \tan^2 x} = \tan x \)
