`-` Đáp án + Giải thích các bước giải:

 `-` a)

`VT = (1 + \tan x)(1 + \cot x) \sin x \cos x` 

$= \left(1 + \frac{\sin x}{\cos x}\right)\left(1 + \frac{\cos x}{\sin x}\right) \sin x \cos x$ 

$= \left(\frac{\cos x + \sin x}{\cos x}\right)\left(\frac{\sin x + \cos x}{\sin x}\right) \sin x \cos x$ 

$= \frac{(\sin x + \cos x)^2}{\sin x \cos x} \sin x \cos x$ 

$= (\sin x + \cos x)^2$ 

$ = \sin^2 x + 2 \sin x \cos x + \cos^2 x$ 

`=1+sin2x=VP`

`-` b)

$VT = \frac{\sin x + \cos x - 1}{1 - \cos x}$ 

$= \frac{(\sin x + \cos x - 1)(\sin x - \cos x + 1)}{(1 - \cos x)(\sin x - \cos x + 1)}$ 

`+)` Tử số:

$(\sin x + \cos x - 1)(\sin x - \cos x + 1) = [\sin x + (\cos x - 1)][\sin x - (\cos x - 1)]$

$= \sin^2 x - (\cos x - 1)^2$

$= \sin^2 x - (\cos^2 x - 2 \cos x + 1)$ 

$= \sin^2 x - \cos^2 x + 2 \cos x - 1$ 

$= (1 - \cos^2 x) - \cos^2 x + 2 \cos x - 1$

$= -2 \cos^2 x + 2 \cos x$

$= 2 \cos x (1 - \cos x)$

 `=>` $VT= \frac{2 \cos x (1 - \cos x)}{(1 - \cos x)(\sin x - \cos x + 1)}$

$= \frac{2 \cos x}{\sin x - \cos x + 1}$