\(a,=\frac{2cos^2\alpha-cos^2\alpha-sin^2\alpha}{sin\alpha+cos\alpha}\\ =\frac{cos^2\alpha-sin^2\alpha}{sin\alpha+cos\alpha}\\ =cos\alpha-sin\alpha\)

\(b,sin25=cos65;cos70=sin20;Khiđó:B=1\)

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 \(\sin^2a+\cos^2a=1\)

=\(\frac{\cos^2a-\sin^2a}{\sin a+\cos a}=\frac{\left(\cos a-\sin a\right)\left(\cos a+\sin a\right)}{\sin a+\cos a}\)

=\(\cos a-\sin a\)

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

\(=2008+1=2009\)

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 \(A< 0,\forall20^0< \alpha< 90^0\)

Bài 1: 

\(=\left(\sin20^0-\cos70^0\right)+\left(-\tan40^0+\cot50^0\right)=0+0=0\)

Bài 2: 

\(\cos a=\sqrt{1-\dfrac{4}{9}}=\dfrac{\sqrt{5}}{3}\)

\(A=2\cdot\sin^2a+3\cdot\cos^2a=2\cdot\dfrac{4}{9}+3\cdot\dfrac{5}{9}=\dfrac{8+15}{9}=\dfrac{23}{9}\)

giúp mình đi!!

\(\sin^225^o+\sin^265^o-\tan35^o+\cot55^o-\frac{\cot32^o}{tan58^o}\)

\(=\cos^265^o+\sin^265^o-\cot55^{^{ }o}+\cot55^o-\frac{\tan58^o}{\tan58^o}\)

\(=1-0-1\)

\(=0\)

nhớ k cho mik nha ^^

\(VT=\dfrac{\sin25^0}{\cos25^0}+2\left(\sin^215^0+\cos^215^0\right)-\tan25^0+4\cdot\dfrac{1}{2}\\ =\tan25^0+2\cdot1-\tan25^0+2=4\)