\(=2008\left(\sin^220^o+\cos^220^o\right)+\cos70^o-\cos70^o+\frac{\sin20^o}{\cos20}.\frac{sin70}{c\text{os}70}\)

\(=2008+1=2009\)

Ta có sin25°=cos65°

         cos70°=20sin°

=> sịn25°+cos70°/sin20°+cos65°=cos65°+sin20°/sin20°+cos65°=1

Sửa: \(A=\dfrac{\cos70^0-\sin\alpha}{\tan60^0-\cot70^0}\)

Vì \(\sin\alpha>\sin20^0\Leftrightarrow\cos70^0-\sin\alpha< \sin20^0-\sin20^0=0\)

Mà \(\tan60^0-\cot70^0=\tan60^0-\tan20^0>0\)

Do đó \(A< 0,\forall20^0< \alpha< 90^0\)

CÁC BN CHỈ CẦN LÀM CHO MIK CÂU D,E,F LÀ ĐC RỒI

d/ \(sin35+sin67-cos23-cos55\)

\(=sin35+sin67-sin67-sin35=0\)

e/ \(cos^220+cos^240+cos^250+cos^270\)

\(=cos^220+cos^240+sin^220+sin^240=1+1=2\)

f/ Đề sai.

câu f sin bộ bình ik nha bn

\(S=\frac{\sqrt{3}sin70-cos70}{\sqrt{3}sin70.cos70}=\frac{\frac{\sqrt{3}}{2}sin70-\frac{1}{2}cos70}{\frac{\sqrt{3}}{4}.2sin70.cos70}\)

\(=\frac{sin70.cos30-sin30.cos70}{\frac{\sqrt{3}}{4}sin140}=\frac{sin\left(70-30\right)}{\frac{\sqrt{3}}{4}sin\left(180-40\right)}\)

\(=\frac{sin40}{\frac{\sqrt{3}}{4}sin40}=\frac{4}{\sqrt{3}}\)

a: \(=\left(\cos^215^0+\cos^275^0\right)+\left(\cos^225^0+\cos^265^0\right)+\left(\cos^235^0+\cos^255^0\right)+\cos^245^0\)

=1+1+1+1/2

=3,5

b: \(=\left(\sin^210^0+\sin^280^0\right)-\left(\sin^220^0+\sin^270^0\right)+\left(\sin^230^0\right)-\left(\sin^240^0+\sin^250^0\right)\)

=1-1-1+1/4

=-1+1/4=-3/4

c: \(=\left(\sin15^0-\cos75^0\right)+\left(\sin75^0-\cos15^0\right)+\sin30^0\)

=1/2

bài 2 là tính tan C nhá

mik vt nhầm

A= cos² 20^o + cos² 30^o + cos² 40^o + sin² 40^o +sin² 30^o +sin²20^o

A = (cos² 20^o +sin²20^o) + ( cos² 30^o+sin² 30^o) +(cos² 40^o + sin² 40^o)

A= 1+1+1=3