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

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

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

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

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

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

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

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) \(A = \cos {0^o} + \cos {40^o} + \cos {120^o} + \cos {140^o}\)

Tra bảng giá trị lượng giác của một số góc đặc biệt, ta có:

 \(\cos {0^o} = 1;\;\cos {120^o} =  - \frac{1}{2}\)

Lại có: \(\cos {140^o} =  - \cos \left( {{{180}^o} - {{40}^o}} \right) =  - \cos {40^o}\)  

\(\begin{array}{l} \Rightarrow A = 1 + \cos {40^o} + \left( { - \frac{1}{2}} \right) - \cos {40^o}\\ \Leftrightarrow A = \frac{1}{2}.\end{array}\)

b) \(B = \sin {5^o} + \sin {150^o} - \sin {175^o} + \sin {180^o}\)

Tra bảng giá trị lượng giác của một số góc đặc biệt, ta có:

 \(\sin {150^o} = \frac{1}{2};\;\sin {180^o} = 0\)

Lại có: \(\sin {175^o} = \sin \left( {{{180}^o} - {{175}^o}} \right) = \sin {5^o}\)  

\(\begin{array}{l} \Rightarrow B = \sin {5^o} + \frac{1}{2} - \sin {5^o} + 0\\ \Leftrightarrow B = \frac{1}{2}.\end{array}\)

c) \(C = \cos {15^o} + \cos {35^o} - \sin {75^o} - \sin {55^o}\)

Ta có: \(\sin {75^o} = \cos\left( {{{90}^o} - {{75}^o}} \right) = \cos {15^o}\); \(\sin {55^o} = \cos\left( {{{90}^o} - {{55}^o}} \right) = \cos {35^o}\)

\(\begin{array}{l} \Rightarrow C = \cos {15^o} + \cos {35^o} - \cos {15^o} - \cos {35^o}\\ \Leftrightarrow C = 0.\end{array}\)

d) \(D = \tan {25^o}.\tan {45^o}.\tan {115^o}\)

Ta có: \(\tan {115^o} =  - \tan \left( {{{180}^o} - {{115}^o}} \right) =  - \tan {65^o}\)

Mà: \(\tan {65^o} = \cot \left( {{{90}^o} - {{65}^o}} \right) = \cot {25^o}\)

\(\begin{array}{l} \Rightarrow D = \tan {25^o}.\tan {45^o}.(-\cot {25^o})\\ \Leftrightarrow D =- \tan {45^o} = -1\end{array}\)

e) \(E = \cot {10^o}.\cot {30^o}.\cot {100^o}\)

Ta có: \(\cot {100^o} =  - \cot \left( {{{180}^o} - {{100}^o}} \right) =  - \cot {80^o}\)

Mà: \(\cot {80^o} = \tan \left( {{{90}^o} - {{80}^o}} \right) = \tan {10^o}\Rightarrow \cot {100^o}  =- \tan {10^o}\)

\(\begin{array}{l} \Rightarrow E = \cot {10^o}.\cot {30^o}.(-\tan {10^o})\\ \Leftrightarrow E = -\cot {30^o} =- \sqrt 3 .\end{array}\)

\(H=cot15^o.cot35^o.cot55^o.cot75^o\)

\(=\left(cot15^o.cot75^o\right).\left(cot35^o.cot55^o\right)\)

\(=\left(cot15^o.tan15^o\right).\left(cot35^o.tan35^o\right)\)

\(=1\)

\(I=tan10^o.tan20^o.tan30^o....tan80^o\)

\(=\left(tan10^o.cot10^o\right).\left(tan20^o.cot20^o\right).\left(tan30^o.cot30^o\right).\left(tan40^o.cot40^o\right)\)

\(=1\)

`sin 65^o - cos 25^o =sin65^o - sin65^o =0`

       `->\bb C`

CHọn C

C