a:Sửa đề: \(I=\frac{cos\left(-288^0\right)\cdot\cot72^0}{\tan\left(-162^0\right)\cdot\sin108^0}-\tan18^0\)

Ta có: \(I=\frac{cos\left(-288^0\right)\cdot\cot72^0}{\tan\left(-162^0\right)\cdot\sin108^0}-\tan18^0\)

\(=\frac{cos\left(72^0-360^0\right)\cdot\cot72^0}{\tan\left(18^0-180^0\right)\cdot\sin108^0}-\tan18^0\)

\(=\frac{cos72^0\cdot\cot72^0}{\tan18^0\cdot\sin108^0}-\tan18^0\)

\(=\frac{cos72^0\cdot\frac{cos72^0}{\sin72^0}}{\tan18^0\cdot\sin108^0}-\tan18^0=\frac{cos^272^0}{\sin72^0\cdot\tan18^0\cdot\sin\left(180^0-108^0\right)}-\tan18^0\)

\(=\frac{cos^272^0}{\sin72^0\cdot\tan18^0\cdot\sin72^0}-\tan18^0=\frac{cos^272^0}{\sin^272^0\cdot\tan18^0}-\tan18^0\)

\(=\frac{cos^272^0}{\sin^272^0\cdot\cot72^0}-\tan18^0=\frac{cos^272^0}{\sin^272^0\cdot\frac{cos72^0}{\sin72^0}}-\tan18^0\)

\(=\frac{cos72^0}{\sin72^0}-\tan18^0=\cot72^0-\tan18^0=0\)

b: Ta có: \(J=2\cdot\sin\left(790^0+x\right)+cos\left(1260^0-x\right)+\tan\left(630^0+x\right)\cdot\tan\left(1260^0-x\right)\)

\(=2\cdot\sin\left(720^0+70^0+x\right)+cos\left(1080^0+180^0-x\right)+\tan\left(720^0+x-90^0\right)\cdot\tan\left(1080^0+180^0-x\right)\)

\(=2\cdot\sin\left(70^0+x\right)+cos\left(180^0-x\right)+\tan\left(x-90^0\right)\cdot\tan\left(180^0-x\right)\)

\(=2\cdot\sin\left(70^0+x\right)-cosx-\tan\left(90^0-x\right)\cdot\left(-\tan x\right)\)

\(=2\cdot\sin\left(70^0+x\right)-cosx+\tan\left(90^0-x\right)\cdot\tan x\)

\(=2\cdot\sin\left(70^0+x\right)-cosx+\cot x\cdot\tan x=2\cdot\sin\left(x+70^0\right)-cosx+1\)

a, \(\cos^215+\cos^225+\cos^235+\cos^245+\sin^235+\sin^225+\sin^215\)

=\(\left(\cos^215+\sin^215\right)+\left(\cos^225+\sin^225\right)+\left(\cos^235+\sin^235\right)+\cos^245\)

=\(1+1+1+\frac{1}{2}=\frac{7}{2}\)

b.\(\sin^210-\sin^220-\sin^230-\sin^240-\cos^240-\cos^220+\cos^210\)

=\(\left(\sin^210+\cos^210\right)-\left(\sin^220+\cos^220\right)-\left(\sin^240+\cos^240\right)-\sin^230\)

=\(1-1-1-\frac{1}{4}=-\frac{5}{4}\)

c,\(\sin15+\sin75-\sin75-\cos15+\sin30=\sin30=\frac{1}{2}\)

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

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

Ta có : \(cos^215^o=sin^275^o;cos^225^o=sin^265^o;cos^235^o=sin^255^o;\frac{cos^245^o}{2}=\frac{sin^245^o}{2}\)

Khi đó \(N=sin^275^o+cos^275^o-\left(sin^265^o+cos^265^o\right)+sin^255^o+cos^255^o-\left(\frac{sin^245^0+cos^245^o}{2}\right)\)

Áp dụng công thức \(sin^2a+cos^2a=1\)ta được 

\(N=1-1+1-\frac{1}{2}=\frac{1}{2}\)

Vậy N = 1/2 

câu b chờ chút mình làm cho nhé <33

Ta có : \(cos^21^o=sin^289^o;cos^22^o=sin^288^o;...;cos^244^o=sin^246^o;\frac{cos^245^o}{2}=\frac{sin^245^o}{2}\)

Khi đó \(A=\frac{sin^245^o+cos^245^o}{2}+\left(sin^246^0+cos^246^o\right)+...+\left(sin^289^o+cos^289^o\right)\)

Áp dụng ct \(sin^2a+cos^2a=1\)ta được \(A=\frac{1}{2}+1+1+...+1=...\)

P/S : bạn tự đếm xem bao nhiêu cặp nhé ;) tìm ssh á 

a)

Đặt  \(A = \left( {2\sin {{30}^o} + \cos {{135}^o} - 3\tan {{150}^o}} \right).\left( {\cos {{180}^o} - \cot {{60}^o}} \right)\)

Ta có: \(\left\{ \begin{array}{l}\cos {135^o} =  - \cos {45^o};\cos {180^o} =  - \cos {0^o}\\\tan {150^o} =  - \tan {30^o}\end{array} \right.\)

\( \Rightarrow A = \left( {2\sin {{30}^o} - \cos {{45}^o} + 3\tan {{30}^o}} \right).\left( { - \cos {0^o} - \cot {{60}^o}} \right)\)

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

\(\left\{ \begin{array}{l}\sin {30^o} = \frac{1}{2};\tan {30^o} = \frac{{\sqrt 3 }}{3}\\\cos {45^o} = \frac{{\sqrt 2 }}{2};\cos {0^o} = 1;\cot {60^o} = \frac{{\sqrt 3 }}{3}\end{array} \right.\)

\( \Rightarrow A = \left( {2.\frac{1}{2} - \frac{{\sqrt 2 }}{2} + 3.\frac{{\sqrt 3 }}{3}} \right).\left( { - 1 - \frac{{\sqrt 3 }}{3}} \right)\)

\(\begin{array}{l} \Leftrightarrow A =  - \left( {1 - \frac{{\sqrt 2 }}{2} + \sqrt 3 } \right).\left( {1 + \frac{{\sqrt 3 }}{3}} \right)\\ \Leftrightarrow A =  - \frac{{2 - \sqrt 2  + 2\sqrt 3 }}{2}.\frac{{3 + \sqrt 3 }}{3}\\ \Leftrightarrow A =  - \frac{{\left( {2 - \sqrt 2  + 2\sqrt 3 } \right)\left( {3 + \sqrt 3 } \right)}}{6}\\ \Leftrightarrow A =  - \frac{{6 + 2\sqrt 3  - 3\sqrt 2  - \sqrt 6  + 6\sqrt 3  + 6}}{6}\\ \Leftrightarrow A =  - \frac{{12 + 8\sqrt 3  - 3\sqrt 2  - \sqrt 6 }}{6}.\end{array}\)

b)

Đặt  \(B = {\sin ^2}{90^o} + {\cos ^2}{120^o} + {\cos ^2}{0^o} - {\tan ^2}60 + {\cot ^2}{135^o}\)

Ta có: \(\left\{ \begin{array}{l}\cos {120^o} =  - \cos {60^o}\\\cot {135^o} =  - \cot {45^o}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{\cos ^2}{120^o} = {\cos ^2}{60^o}\\{\cot ^2}{135^o} = {\cot ^2}{45^o}\end{array} \right.\)

\( \Rightarrow B = {\sin ^2}{90^o} + {\cos ^2}{60^o} + {\cos ^2}{0^o} - {\tan ^2}60 + {\cot ^2}{45^o}\)

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

\(\left\{ \begin{array}{l}\cos {0^o} = 1;\;\;\cot {45^o} = 1;\;\;\cos {60^o} = \frac{1}{2}\\\tan {60^o} = \sqrt 3 ;\;\;\sin {90^o} = 1\end{array} \right.\)

\( \Rightarrow B = {1^2} + {\left( {\frac{1}{2}} \right)^2} + {1^2} - {\left( {\sqrt 3 } \right)^2} + {1^2}\)

\( \Leftrightarrow B = 1 + \frac{1}{4} + 1 - 3 + 1 = \frac{1}{4}.\)

c

Đặt  \(C = \cos {60^o}.\sin {30^o} + {\cos ^2}{30^o}\)

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

\(\sin {30^o} = \frac{1}{2};\;\;\cos {30^o} = \frac{{\sqrt 3 }}{2};\;\cos {60^o} = \frac{1}{2}\;\)

\( \Rightarrow C = \frac{1}{2}.\frac{1}{2} + {\left( {\;\frac{{\sqrt 3 }}{2}} \right)^2} = \frac{1}{4} + \frac{3}{4} = 1.\)

\(a,A=\sin^234^0+\cos^234^0+\dfrac{\cot42^0}{\cot42^0}=1+1=2\\ b,B=\left(\cos^213^0+\sin^277^0\right)+\dfrac{3\cot64^0}{\cot64^0}+2\cot32^0\cdot\tan32^0\\ B=1+3+2\cdot1=6\\ c,B=\dfrac{5\cot35^0}{\cot35^0}-2\left(\sin^261^0-\cos^261^0\right)=5-2\cdot1=3\)

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

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

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

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

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

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

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