Đáp án: $\dfrac{\tan^2x+2\tan x+2}{\tan^2x}$
Giải thích các bước giải:
Ta có :
$\dfrac{\tan x+1-2\cot^2x}{\tan x-1}$
$=\dfrac{\tan^3x+\tan^2x-2\tan^2x\cot^2x}{\tan^2x(\tan x-1)}$
$=\dfrac{\tan^3x+\tan^2x-2}{\tan^2x(\tan x-1)}$
$=\dfrac{\tan^3x-1+\tan^2x-1}{\tan^2x(\tan x-1)}$
$=\dfrac{(\tan x-1)(\tan^2x+\tan x+1)+(\tan x-1)(\tan x+1)}{\tan^2x(\tan x-1)}$
$=\dfrac{(\tan x-1)(\tan^2x+\tan x+1+\tan x+1)}{\tan^2x(\tan x-1)}$
$=\dfrac{(\tan x-1)(\tan^2x+2\tan x+2)}{\tan^2x(\tan x-1)}$
$=\dfrac{\tan^2x+2\tan x+2}{\tan^2x}$
