a: 

b: \(B=3-sin^290^0+2\cdot cos^260^0-3\cdot tan^245^0\)

\(=3-1+2\cdot\left(\dfrac{1}{2}\right)^2-3\cdot1^2\)

\(=2-3+2\cdot\dfrac{1}{4}=-1+\dfrac{1}{2}=-\dfrac{1}{2}\)

c: \(C=sin^245^0-2\cdot sin^250^0+3\cdot cos^245^0-2\cdot sin^240^0+4\cdot tan55\cdot tan35\)

\(=\left(\dfrac{\sqrt{2}}{2}\right)^2+3\cdot\left(\dfrac{\sqrt{2}}{2}\right)^2-2\cdot\left(sin^250^0+sin^240^0\right)+4\)

\(=\dfrac{1}{2}+3\cdot\dfrac{1}{2}-2+4\)

\(=2-2+4=4\)

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

\(a,\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{AC}{BC}:\dfrac{AB}{BC}=\dfrac{AC}{AB}=\tan\alpha\\ \cot\alpha=\dfrac{1}{\tan\alpha}=\dfrac{1}{\dfrac{\sin\alpha}{\cos\alpha}}=\dfrac{\cos\alpha}{\sin\alpha}\\ \tan\alpha\cdot\cot\alpha=\dfrac{AC}{AB}\cdot\dfrac{AB}{AC}=1\\ b,\sin^2\alpha+\cos^2\alpha=\dfrac{AC^2}{BC^2}+\dfrac{AB^2}{BC^2}=\dfrac{AB^2+AC^2}{BC^2}=\dfrac{BC^2}{BC^2}=1\left(định.lí.pytago\right)\)

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

=>\(tan20< tan40< tan53< tan82\)

=>\(tan20< tan40< cot37< cot8\)

=>\(sin20< tan20< tan40< cot37< cot8\)

Ta có: \(\sin {70^o} = \cos {20^o};\;\cos {110^o} =  - \cos {70^o} =  - \sin {20^o}\)

\(\begin{array}{l} \Rightarrow A = {(\sin {20^o} + \cos {20^o})^2} + {(\cos {20^o} - \sin {20^o})^2}\\ = ({\sin ^2}{20^o} + {\cos ^2}{20^o} + 2\sin {20^o}\cos {20^o}) + ({\cos ^2}{20^o} + {\sin ^2}{20^o} - 2\sin {20^o}\cos {20^o})\\ = 2({\sin ^2}{20^o} + {\cos ^2}{20^o})\\ = 2\end{array}\)

Ta có: \(\tan {110^o} =  - \tan {70^o} =  - \cot {20^o};\;\cot {110^o} =  - \cot {70^o} =  - \tan {20^o}.\)

\( \Rightarrow B = \tan {20^o} + \cot {20^o} + ( - \cot {20^o}) + ( - \tan {20^o}) = 0\)

a/

\(tana+tanb=\frac{sina}{cosa}+\frac{sinb}{cosb}=\frac{sinacosb+cosa.sinb}{cosa.cosb}=\frac{sin\left(a+b\right)}{cosa.cosb}\)

\(C=tan80\left(tan20+tan140\right)+tan20.tan120\)

\(C=tan80.\frac{sin160}{cos20.cos140}+\frac{sin20.sin140}{cos20.cos140}\)

\(C=\frac{sin80}{cos80}.\frac{2.sin80.cos80}{\frac{1}{2}\left(cos160+cos120\right)}+\frac{-\frac{1}{2}\left(cos160-cos120\right)}{\frac{1}{2}\left(cos160+cos120\right)}\)

\(C=\frac{4sin^280}{cos160+cos120}-\frac{cos160-cos120}{cos160+cos120}\)

\(C=\frac{2\left(1-cos160\right)-cos160+cos120}{cos160+cos120}=\frac{2+cos120-3cos160}{cos120+cos160}\)

\(C=\frac{2-\frac{1}{2}-3cos160}{-\frac{1}{2}+cos160}=\frac{3-6cos160}{2cos160-1}=-3\)

b/

\(cos^275-sin^275=cos150=-\frac{\sqrt{3}}{2}\)