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Bài 1:
\(\cos60^0=\sin30^0;\sin67^0=\cos23^0;\tan80^0=\cot10^0;\cot20^0=\cot20^0\)
Bài 2:
Xét tam giác ABC vuông tại A
\(a,\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{AC}{BC}:\dfrac{AB}{BC}=\dfrac{AC}{AB}=\tan\alpha\\ \cot\alpha=\dfrac{1}{\tan\alpha}=\dfrac{1}{\dfrac{\sin\alpha}{\cos\alpha}}=\dfrac{\cos\alpha}{\sin\alpha}\\ \tan\alpha\cdot\cot\alpha=\dfrac{AC}{AB}\cdot\dfrac{AB}{AC}=1\\ b,\sin^2\alpha+\cos^2\alpha=\dfrac{AC^2}{BC^2}+\dfrac{AB^2}{BC^2}=\dfrac{AB^2+AC^2}{BC^2}=\dfrac{BC^2}{BC^2}=1\left(định.lí.pytago\right)\)
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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
=>\(tan20< tan40< tan53< tan82\)
=>\(tan20< tan40< cot37< cot8\)
=>\(sin20< tan20< tan40< cot37< cot8\)
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A=(sin220°+sin270°)+(sin230°+sin260°)
+(sin240°+sin250°)-tan245°
=(sin220°+cos220°)+(sin230°+cos230°)+(sin240°+cos240°)-1
=1+1+1-1=2
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