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\(a,=\frac{2cos^2\alpha-cos^2\alpha-sin^2\alpha}{sin\alpha+cos\alpha}\\ =\frac{cos^2\alpha-sin^2\alpha}{sin\alpha+cos\alpha}\\ =cos\alpha-sin\alpha\)
\(b,sin25=cos65;cos70=sin20;Khiđó:B=1\)
b) \(\frac{\sin25+\cos70}{\sin20+\cos65}\)
xét tam giác vuông có : sin a= cos b => cos 70 = sin (90 -70) <=> cos 70 = sin 20
cos 65 =sin 25
<=> \(\frac{\sin25+\cos70}{\sin20+\cos65}\)
=\(\frac{\sin25+\sin20}{\sin20+\sin25}=1\)
\(\frac{2\cos^2\cdot a-1}{\sin a+\cos a}=\frac{2\cos^2a-\left(\sin^2+\cos^2\right)}{\sin a+\cos a}\)
vì \(\sin^2a+\cos^2a=1\)
=\(\frac{\cos^2a-\sin^2a}{\sin a+\cos a}=\frac{\left(\cos a-\sin a\right)\left(\cos a+\sin a\right)}{\sin a+\cos a}\)
=\(\cos a-\sin a\)
\(=2008\left(\sin^220^o+\cos^220^o\right)+\cos70^o-\cos70^o+\frac{\sin20^o}{\cos20}.\frac{sin70}{c\text{os}70}\)
\(=2008+1=2009\)
\(\sin^225^o+\sin^265^o-\tan35^o+\cot55^o-\frac{\cot32^o}{tan58^o}\)
\(=\cos^265^o+\sin^265^o-\cot55^{^{ }o}+\cot55^o-\frac{\tan58^o}{\tan58^o}\)
\(=1-0-1\)
\(=0\)
nhớ k cho mik nha ^^
Lời giải:
a)
\(\frac{\cos 65}{\sin 25}=\frac{\cos (90-25)}{\sin 25}=\frac{\sin 25}{\sin 25}=1\)
\(\cot 35-\cot 55=\cot 35-\cot (90-35)=\cot 35-\tan 35\)
\(=\frac{\cos 35}{\sin 35}-\frac{\sin 35}{\cos 35}=\frac{\cos ^235-\sin ^235}{\sin 35.\cos 35}=\frac{\cos (2.35)}{\sin 35.\cos 35}=\frac{2\cos 70}{2\sin 35\cos 35}=\frac{2\cos 70}{\sin (2.35)}\)
\(=\frac{2\cos 70}{\sin 70}=2\cot 70\)
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 \(A< 0,\forall20^0< \alpha< 90^0\)
Ta có sin25°=cos65°
cos70°=20sin°
=> sịn25°+cos70°/sin20°+cos65°=cos65°+sin20°/sin20°+cos65°=1