1)Bạn tự vẽ hình dùm mình :(

sinB = sin48độ = $\frac{AC}{BC}$ = xấp xỉ 0,7431

->BC = xấp xỉ 26,91 (cm)

Góc C = 90 - 48 = 42 độ

SinC = Sin 42 độ = $\frac{AB}{BC}$ = xấp xỉ 0,6691

-> AB = xấp xỉ 18 (cm)

2. $\frac{sin}{1+cos}$ + $\frac{1+cos}{sin}$ 

= $\frac{sin^2+(1+cos)^2}{(1+cos)sin}$ 

= $\frac{sin^2+1+2cos+cos^2}{sin(1+cos)}$ 

= $\frac{1+1+2cos}{sin(1+cos)}$

= $\frac{2(1+cos)}{sin(1+cos)}$ 

= $\frac{2}{sin}$  (đpcm)

3) $\frac{1-cos}{sin}$ - $\frac{sin}{1+cos}$ 

= $\frac{(1-cos)(1+cos)-sin^2}{sin(1+cos)}$ 

= $\frac{1-cos^2-sin^2}{sin(1 +cos)}$ 

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

= 0 (đpcm)

4) $cos^{4}+cos^2sin^2+sin^2$ 

= $cos^{2}(cos^2+sin^2)+sin^2$ 

= $cos^{2}.1+sin^2$ 

= $cos^{2}+sin^2$ 

= 1 (đpcm)

Chúc bạn học tốt !!!!!!

`1)` $AC=20cm; \hat{B}=48°$

$∆ABC$ vuông tại $A$

`=>cotB={AB}/{AC}`

`=>AB=AC.cotB=20.cot48°≈18cm`

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`\qquad sinB={AC}/{BC}`

`=>BC={AC}/{sin}={20}/{sin48°}≈27cm`

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`\qquad \hat{B}+\hat{C}=90°` (hai góc phụ nhau)

`=>\hat{C}=90°-\hat{B}=90°-48°=42°`

Vậy: $AB≈18cm; BC≈27cm;\hat{C}=42°$

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`2)` Tính chất:

`sin^2 x +cos^2x=1`

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Ta có:

`\qquad {sinx}/{1+cosx}+{1+cosx}/{sinx}`

`={sin^2x+(1+cosx)^2}/{(1+cosx).sinx}`

`={sin^2 x+1+2cosx+cos^2x}/{(1+cosx).sinx}`

`={sin^2 x +cos^2x+1+2cosx}/{(1+cosx).sinx}`

`={1+1+2cosx}/{(1+cosx).sinx}`

`={2(1+cosx)}/{(1+cosx).sinx}=2/{sinx}`

Vậy: `{sinx}/{1+cosx}+{1+cosx}/{sinx}=2/{sinx}` (đpcm)

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`3)` `{1-cosx}/{sinx}-{sinx}/{1+cosx}`

`={(1-cosx)(1+cosx)-sin^2x}/{sinx(1+cosx)}`

`={1-cos^2x-sin^2x}/{sinx(1+cosx)}`

`={1-(sin^2x+cos^2 x)}/{sinx(1+cosx)}`

`={1-1}/{sinx(1+cosx)}=0`

Vậy `{1-cosx}/{sinx}-{sinx}/{1+cosx}=0\ (đpcm)`

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`4)` `cos^4 x+sin^2 x cos^2 x +sin^2 x`

`=cos^2 (cos^2 x+sin^2 x)+sin^2 x`

`=cos^2 x .1+sin^2 x=1`

Vậy `cos^4 x+sin^2 x cos^2 x +sin^2 x=1\ (đpcm)`