1) \(\cot51^0=\tan39^0\)

\(\cot79^015'=\tan10^045'\)

Do đó: \(\cot79^015'< \tan13^0< \tan28^0< \cot51^0< \tan47^0\)

2) \(\cos62^0=\sin28^0\)

\(\cos63^041'=\sin26^019'\)

\(\cos87^0=\sin3^0\)

Do đó: \(\cos87^0< \cos63^041'< \cos62^0< \sin47^0< \sin50^0\)

a,

Đổi `tan 12^o = cot 78^o ; tan 28^o = cot 62^o ; tan 58^o = cot 32^o`

Vì `32^o<61^o<62^o<78^o<79^15'`

`->cot 32^o>cot 61^o>cot 62^o > cot 78^o > cot 79^o15'`

`->tan 58^o>cot 61^o > tan 28^o > tan 12^o > cot 79^o15'`

b,

Đổi `sin 56^o = cos 34^o ; sin 74^o=cos 16^o`

Vì `16^o<24^o<63^o41'<67^o<85 ^o`

`->cos 16^o>cos 34^o>cos 63^o41'>cos 67^o>cos 85 ^o`

`->sin 74^o>sin 56^o>cos 63^o41'>cos 67^o>cos 85 ^o`

a: \(cos32=sin58;cos53=sin37;cos8=sin82\)

18<37<44<58<82

=>\(sin18< sin37< sin44< sin58< sin82\)

=>\(sin18< cos53< sin44< cos32< cos8\)

b: 20<45

=>\(sin20< tan20\)

\(cot8=tan82;cot37=tan53\)

20<40<53<82

=>\(tan20< tan40< tan53< tan82\)

=>\(tan20< tan40< cot37< cot8\)

=>\(sin20< tan20< tan40< cot37< cot8\)

Thứ tự tăng dần :

1) cos 62 độ 25 phút; sin 35 độ; cos 47 độ; sin 53 độ 30 phút; sin 74 độ.

2) tan 11 độ; cot 63 độ = tan 27 độ; cot 57 độ 30 phút; tan 55 độ

\(A=sin42^0-cos48^0=cos\left(90^0-42^0\right)-cos48^0=cos48^0-cos48^0=0\)

\(B=cot56^0-tan34^0=tan\left(90^0-56^0\right)-tan34^0=tan34^0-tan34^0=0\)

\(C=sin30^0-cot50^0-cos60^0+tan40^0\)

\(=cos\left(90^0-30^0\right)-tan\left(90^0-50^0\right)-cos60^0+tan40^0\)

\(=cos60^0-tan40^0-cos60^0+tan40^0=0\)

\(A=\sin42^0-\cos48^0=\sin42^0-\sin42^0=0\)

\(B=\cot56^0-\tan34^0=\tan34^0-\tan34^0=0\)

1: 

a: sin a=căn 3/2

\(cosa=\sqrt{1-sin^2a}=\sqrt{1-\dfrac{3}{4}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)

\(tana=\dfrac{\sqrt{3}}{2}:\dfrac{1}{2}=\sqrt{3}\)

cot a=1/tan a=1/căn 3

b: \(tana=2\)

=>cot a=1/tan a=1/2

\(1+tan^2a=\dfrac{1}{cos^2a}\)

=>\(\dfrac{1}{cos^2a}=5\)

=>cos^2a=1/5

=>cosa=1/căn 5

\(sina=\sqrt{1-cos^2a}=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}\)

c: \(cosa=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)

tan a=5/13:12/13=5/12

cot a=1:5/12=12/5