`~rai~`

\(3.\tan^2x-\sin^2x=\tan^2x\sin^2x\\\text{Xét VT=}\tan^2x-\sin^2x\\\quad\quad\quad=\dfrac{\sin^2x}{\cos^2x}-\sin^2x\\\quad\quad\quad=\dfrac{\sin^2x-\sin^2x\cos^2x}{\cos^2x}\\\quad\quad\quad=\dfrac{\sin^2x(1-\cos^2x)}{\cos^2x}\\\quad\quad\quad=\dfrac{\sin^2x.\sin^2x}{\cos^2x}\\\quad\quad\quad=\dfrac{\sin^2x}{\cos^2x}.\sin^2x\\\quad\quad\quad=\tan^2x\sin^2x\\\quad\quad\quad=VP.\\\Rightarrow \tan^2x-\sin^2x=\tan^2x\sin^2x.(đpcm)\\4.1+\sin x+\cos x+\tan x=(1+\cos x)(1+\tan x)\\\text{Xét VP=}(1+\cos x)(1+\tan x)\\\quad\quad\quad=1+\tan x+\cos x+\cos x.\tan x\\\quad\quad\quad=1+\tan x+\cos x+\cos x.\dfrac{\sin x}{\cos x}\\\quad\quad\quad=1+\tan x+\cos x+\sin x\\\quad\quad\quad=1+\sin x+\cos x+\tan x\\\quad\quad\quad=VT.\\\Rightarrow 1+\sin x+\cos x+\tan x=(1+\cos x)(1+\tan x).(đpcm)\)

3.

`tan^2x-sin^2x=tan^2xsin^2x`

`VT=tan^2x-sin^2x`

`=sin^2x/cos^2x-sin^2x`

`=\frac{sin^2x-sin^2x.cos^2x}{cos^2x}`

`=\frac{sin^2x(1-cos^2x)}{cos^2x}`

`=tan^2x(1-cos^2x)`

`=tan^2xsin^2x=VP`

Vậy `tan^2x-sin^2x=tan^2xsin^2x` $\text{(đpcm)}$

4.

`1+sinx+cosx+tanx=(1+cosx)(1+tanx)`

`VT=1+sinx+cosx+tanx`

`=1+cosx.sinx/cosx+cosx+tanx`

`=1+tanx.cosx+cosx+tanx`

`=(1+tanx)+cosx(1+tanx)`

`=(1+cosx)(1+tanx)=VP`

Vậy `1+sinx+cosx+tanx=(1+cosx)(1+tanx)`  $\text{(đpcm)}$