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

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

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

Lớp 9 nên coi như các góc này đều nhọn

a.

\(cosa=\sqrt{1-sin^2a}=\dfrac{15}{17}\)

\(tana=\dfrac{sina}{cosa}=\dfrac{8}{15}\)

\(cota=\dfrac{1}{tana}=\dfrac{15}{8}\)

b.

\(1+cot^2a=\dfrac{1}{sin^2a}\Rightarrow sina=\dfrac{1}{\sqrt{1+cot^2a}}=\dfrac{4}{5}\)

\(cosa=\sqrt{1-sin^2a}=\dfrac{3}{5}\)

\(tana=\dfrac{1}{cota}=\dfrac{4}{3}\)

a) \(\cos=\sqrt{1-\sin^2}=\sqrt{1-\dfrac{64}{289}}=\dfrac{15}{17}\)

\(\tan=\dfrac{\sin}{\cos}=\dfrac{8}{17}:\dfrac{15}{17}=\dfrac{8}{15}\)

\(\cot=\dfrac{\cos}{\sin}=\dfrac{15}{17}:\dfrac{8}{17}=\dfrac{15}{8}\)

\(tana-cota=2\sqrt{3}\Rightarrow\left(tana-cota\right)^2=12\)

\(\Rightarrow\left(tana+cota\right)^2-4=12\Rightarrow\left(tana+cota\right)^2=16\)

\(\Rightarrow P=4\)

\(sinx+cosx=\dfrac{1}{5}\Rightarrow\left(sinx+cosx\right)^2=\dfrac{1}{25}\)

\(\Rightarrow1+2sinx.cosx=\dfrac{1}{25}\Rightarrow sinx.cosx=-\dfrac{12}{25}\)

\(P=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}=\dfrac{sin^2x+cos^2x}{sinx.cosx}=\dfrac{1}{sinx.cosx}=\dfrac{1}{-\dfrac{12}{25}}=-\dfrac{25}{12}\)

bài này không có giới hạn góc sao tìm được bạn .

\(sin\) \(a\) \(=\dfrac{\sqrt{3}}{2}\Rightarrow\widehat{a}=60^o\)

\(\Rightarrow cos\) \(a\) \(=cot\) \(60^o\) \(=\dfrac{1}{2}\)

\(tan\) \(a\) \(=tan\) \(60^o\) \(=\sqrt{3}\)

\(cot\) \(a\) \(\dfrac{1}{tana}\) \(=\dfrac{1}{\sqrt{3}}\) \(=\dfrac{\sqrt{3}}{\sqrt{3}}\) 

\(\sin a=\frac{\sqrt{3}}{2}\Rightarrow\widehat{a}=60^0\)

\(\Rightarrow\cos a=\cos60^0=\frac{1}{2}\)

    \(\tan a=\tan60^0=\sqrt{3}\)

     \(\cot a=\frac{1}{\tan a}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)

Trả lời : 

sin a=√3/2 ⇒â=60°

⇒cos a=cos 60° =1/2 

    tan a=tan 60°=√3

     cot a=1/tan a =1/√3 =√3/3 

Cho cos a = 1/3