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

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

1.

\(cos70+cos50=2cos\dfrac{70+50}{2}.cos\dfrac{70-50}{2}=2.cos60.cos10=2.\dfrac{1}{2}cos10\)

\(cos70+cos50-cos10=0\)

2.\(tan\left(a+b\right)=\dfrac{tana+tanb}{1-tana.tanb}=1\Rightarrow tana+tanb+tana.tanb+1=2\Leftrightarrow\left(1+tana\right)\left(1+tanb\right)=2\)

Ta có sin25°=cos65°

         cos70°=20sin°

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

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

\(=2008+1=2009\)

Bài 1: 

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

Bài 2: 

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

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

Biểu thức này chỉ rút gọn được khi mẫu là \(1-2sin^210^0\)

\(tan40+tan50=\dfrac{sin40}{cos40}+\dfrac{sin50}{cos50}=\dfrac{sin40.cos50+cos50.sin40}{cos40.cos50}\)

\(=\dfrac{sin\left(40+50\right)}{\dfrac{1}{2}\left(cos90+cos10\right)}=\dfrac{2}{cos10}\)

\(\Rightarrow tan30+tan60+tan40+tan50=\dfrac{\sqrt{3}}{3}+\sqrt{3}+\dfrac{2}{cos10}\)

\(=\dfrac{4\sqrt{3}}{3}+\dfrac{2}{cos10}=\dfrac{4\sqrt{3}cos10+6}{3.cos10}=\dfrac{4\sqrt{3}\left(cos10+\dfrac{\sqrt{3}}{2}\right)}{3.cos10}\)

\(=\dfrac{4\sqrt{3}\left(cos10+cos30\right)}{3cos10}=\dfrac{8\sqrt{3}cos20.cos10}{3cos10}=\dfrac{8\sqrt{3}}{3}cos20\)

\(\Rightarrow G=\dfrac{\dfrac{8\sqrt{3}}{3}cos20}{1-2sin^210}=\dfrac{\dfrac{8\sqrt{3}}{3}cos20}{cos20}=\dfrac{8\sqrt{3}}{3}\)

a) Ta có:  \(\left\{ \begin{array}{l}\sin {100^o} = \sin \left( {{{180}^o} - {{80}^o}} \right) = \sin {80^o}\\\cos {164^o} = \cos \left( {{{180}^o} - {{16}^o}} \right) =  - \cos {16^o}\end{array} \right.\)

\( \Rightarrow \sin {100^o} + \sin {80^o} + \cos {16^o} + \cos {164^o}\)\( = \sin {80^o} + \sin {80^o} + \cos {16^o}-\cos {16^o}\)\( = 2\sin {80^o}.\)

b) 

Ta có:

\(\left\{ \begin{array}{l}\sin \left( {{{180}^o} - \alpha } \right) = \sin \alpha \\\cos \left( {{{180}^o} - \alpha } \right) =  - \cos \alpha \\\tan \left( {{{180}^o} - \alpha } \right) =  - \tan \alpha \\\cot \left( {{{180}^o} - \alpha } \right) =  - \cot \alpha \end{array} \right.\quad ({0^o} < \alpha  < {90^o})\)\( \Rightarrow 2\sin \left( {{{180}^o} - \alpha } \right).\cot \alpha  - \cos \left( {{{180}^o} - \alpha } \right).\tan \alpha .\cot \left( {{{180}^o} - \alpha } \right)\) \( = 2\sin \alpha .\cot \alpha  - \left( { - \cos \alpha } \right).\tan \alpha .\left( { - \cot \alpha } \right)\)\( = 2\sin \alpha .\cot \alpha  - \cos \alpha .\tan \alpha .\cot \alpha \)

\( = 2\sin \alpha .\frac{{\cos \alpha }}{{\sin \alpha }} - \cos \alpha .\left( {\tan \alpha .\cot \alpha } \right)\)\( = 2\cos \alpha  - \cos \alpha .1 = \cos \alpha .\)