1: 

a: sin a=căn 3/2

\(cosa=\sqrt{1-sin^2a}=\sqrt{1-\dfrac{3}{4}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)

\(tana=\dfrac{\sqrt{3}}{2}:\dfrac{1}{2}=\sqrt{3}\)

cot a=1/tan a=1/căn 3

b: \(tana=2\)

=>cot a=1/tan a=1/2

\(1+tan^2a=\dfrac{1}{cos^2a}\)

=>\(\dfrac{1}{cos^2a}=5\)

=>cos^2a=1/5

=>cosa=1/căn 5

\(sina=\sqrt{1-cos^2a}=\sqrt{\dfrac{4}{5}}=\dfrac{2}{\sqrt{5}}\)

c: \(cosa=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)

tan a=5/13:12/13=5/12

cot a=1:5/12=12/5

Ta có: \(cot\alpha=\dfrac{cos\alpha}{sin\alpha}=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}\)

Lại có: \(\dfrac{1}{cot\alpha}=tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{sin^2\alpha}{cos\alpha.sin\alpha}=\dfrac{1}{\sqrt{5}}\)

\(\Rightarrow A=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}+\dfrac{sin^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}+\dfrac{1}{\sqrt{5}}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)

Ta có : cot α = \(\sqrt{5}\Rightarrow\dfrac{cos\alpha}{sin\alpha}=\sqrt{5}\Rightarrow cos\alpha=\sqrt{5}.sin\alpha\)

\(A=\dfrac{sin^2\alpha+cos^2\alpha}{sin\alpha.cos\alpha}\)

\(A=\dfrac{sin^2\alpha+\left(\sqrt{5}sin\alpha\right)^2}{sin\alpha.\sqrt{5}sin\alpha}=\dfrac{sin^2\alpha+5sin^2\alpha}{\sqrt{5}sin^2\alpha}\)

\(A=\dfrac{6sin^2\alpha}{\sqrt{5}sin^2\alpha}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)

a: sin a=2/3

=>cos^2a=1-(2/3)^2=5/9

=>\(cosa=\dfrac{\sqrt{5}}{3}\)

\(tana=\dfrac{2}{3}:\dfrac{\sqrt{5}}{3}=\dfrac{2}{\sqrt{5}}\)

\(cota=1:\dfrac{2}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)

b: cos a=1/5

=>sin^2a=1-(1/5)^2=24/25

=>\(sina=\dfrac{2\sqrt{6}}{5}\)

\(tana=\dfrac{2\sqrt{6}}{5}:\dfrac{1}{5}=2\sqrt{6}\)

\(cota=\dfrac{1}{2\sqrt{6}}=\dfrac{\sqrt{6}}{12}\)

c: cot a=1/tana=1/2

\(1+tan^2a=\dfrac{1}{cos^2a}\)

=>1/cos^2a=1+4=5

=>cos^2a=1/5

=>cosa=1/căn 5

\(sina=\sqrt{1-cos^2a}=\dfrac{2}{\sqrt{5}}\)

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) \(M = \sin {45^o}.\cos {45^o} + \sin {30^o}\)

Ta có: \(\left\{ \begin{array}{l}\sin {45^o} = \cos {45^o} = \frac{{\sqrt 2 }}{2};\;\\\sin {30^o} = \frac{1}{2}\end{array} \right.\)

Thay vào M, ta được: \(M = \frac{{\sqrt 2 }}{2}.\frac{{\sqrt 2 }}{2} + \frac{1}{2} = \frac{2}{4} + \frac{1}{2} = 1\)

b) \(N = \sin {60^o}.\cos {30^o} + \frac{1}{2}.\sin {45^o}.\cos {45^o}\)

Ta có: \(\sin {60^o} = \frac{{\sqrt 3 }}{2};\;\;\cos {30^o} = \frac{{\sqrt 3 }}{2};\;\sin {45^o} = \frac{{\sqrt 2 }}{2};\, \cos {45^o}= \frac{{\sqrt 2 }}{2}\)

Thay vào N, ta được: \(N = \frac{{\sqrt 3 }}{2}.\frac{{\sqrt 3 }}{2} + \frac{1}{2}.\frac{{\sqrt 2 }}{2}.\frac{{\sqrt 2 }}{2} = \frac{3}{4} + \frac{1}{4} = 1\)

c) \(P = 1 + {\tan ^2}{60^o}\)

Ta có: \(\tan {60^o} = \sqrt 3 \)

Thay vào P, ta được: \(Q = 1 + {\left( {\sqrt 3 } \right)^2} = 4.\)

d) \(Q = \frac{1}{{{{\sin }^2}{{120}^o}}} - {\cot ^2}{120^o}.\)

Ta có: \(\sin {120^o} = \frac{{\sqrt 3 }}{2};\;\;\cot {120^o} = \frac{{ - 1}}{{\sqrt 3 }}\)

Thay vào P, ta được: \(Q = \frac{1}{{{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}} - \;{\left( {\frac{{ - 1}}{{\sqrt 3 }}} \right)^2} = \frac{1}{{\frac{3}{4}}} - \;\frac{1}{3} = \;\frac{4}{3} - \;\frac{1}{3} = 1.\)

b) \(\sin x+\cos x=\dfrac{3}{2}\)

\(\left(\sin x+\cos x\right)^2=\dfrac{1}{4}\)

\(\sin^2x+\cos^2x+2\sin x\cos x=\dfrac{1}{4}\)

\(2\sin x\cos x=-\dfrac{3}{4}=\sin2x\)

ý a,

ý c