b) tan a.cot a=1
\(\dfrac{AC}{AB}.\dfrac{AB}{AC}=1\)

bài 3
a) \(tana=\dfrac{sina}{cosa}\)
\(\dfrac{AC}{\dfrac{BC}{\dfrac{AB}{BC}}}=\dfrac{AC}{AB}=tana\)

sin 65⁰=cos 35⁰
\(cos70^0=sin30^0\)
\(tan80^0=cot20^0\)
\(cot68^0=tan32^0\)

có tan x + cot x=5
<=>     2(tan x + cot x) =2.5
<=>      tan 2x +cot 2x =10
có tan x + cot x=5
<=>     3(tan x + cot x) =3.5
<=>      tan 3x +cot 3x =15

góc AOD=góc BOC=60⁰(đối đỉnh)
góc AOC+góc BOC=180⁰
=>     góc AOC+60⁰=180⁰
=>     góc AOC=120⁰
=>     góc BOD=góc AOC=120⁰ (đối đỉnh)

bài 7
\(x^{n+2}-x^n\)
\(=x^{n+2-n}=x^2\)
\(\left(b\right)x^{x+3}-x^{x+1}=x^{x+3-x-1}=X^2\)
c)
\(x^{2m}+x^m=x^{2m+m}=x^{2m}\)
d)
\(x^{2n+1}-x^{4n}=x^{2n+1-4n}=x^{1-2n}\)

1,drivers,went
2,did you talk
3, is staying
4,will come
5.have never eaten
6,didn't you go
7,will play
8,went
9,carving,making
10,cooking,doing

\(16-8x+x^2\)
\(=4^2-2.4.x+x^2\)
\(=\left(4-x\right)^2\)