a) \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-sin^2x=1-\frac{3}{4}=\frac{1}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}cosx=\frac{1}{2}\\cosx=-\frac{1}{2}\end{cases}}\)

- \(cosx=\frac{1}{2}\): 

\(tanx=\frac{sinx}{cosx}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\)

\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)

- \(cosx=\frac{-1}{2}\): 

\(tanx=\frac{sinx}{cosx}=\frac{\frac{\sqrt{3}}{2}}{\frac{-1}{2}}=-\sqrt{3}\)

\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{-\sqrt{3}}=\frac{-\sqrt{3}}{3}\)

b) Bạn làm tương tự câu a) nha. 

\(tana-5cota+4=0\Rightarrow tana-\dfrac{5}{tana}+4=0\)

\(\Rightarrow tan^2a+4tana-5=0\Rightarrow\left[{}\begin{matrix}tana=1\\tana=-5\end{matrix}\right.\)

\(A=\dfrac{4sina+2cosa}{3sina-cosa}=\dfrac{\dfrac{4sina}{cosa}+\dfrac{2cosa}{cosa}}{\dfrac{3sina}{cosa}-\dfrac{cosa}{cosa}}=\dfrac{4tana+2}{3tana-1}=\left[{}\begin{matrix}3\\\dfrac{9}{8}\end{matrix}\right.\)

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

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

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

Ta có

\(\begin{array}{l}Q = {\tan ^2}\frac{\pi }{3} + {\sin ^2}\frac{\pi }{4} + \cot \frac{\pi }{4} + \cos \frac{\pi }{2}\\\,\,\,\,\, = \,{\left( {\sqrt 3 } \right)^2} + {\left( {\frac{{\sqrt 2 }}{2}} \right)^2} + 1 + 0 = \frac{7}{2}\end{array}\)

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

1+\(cot^2\)x=\(\dfrac{1}{sin^2x}\)\(\Leftrightarrow\)1+\(\dfrac{3}{4}^2\)=\(\dfrac{1}{sin^2x}\)➩\(sin^2x\)=\(\dfrac{16}{25}\)\(\Rightarrow\)sinx=\(\dfrac{4}{5}\)

\(\sin^2x+\cos^2x=1\)\(\Rightarrow\)cosx=\(\sqrt{1-sin^2x}\)=\(\dfrac{3}{5}\)

tanx=\(\dfrac{\sin x}{\cos x}\)=\(\dfrac{4}{3}\)

a) \(\frac{1}{cos^2x}=1+tan^2x=1+\frac{9}{16}=\frac{25}{16}\)

\(\Leftrightarrow cos^2x=\frac{16}{25}\Leftrightarrow\orbr{\begin{cases}cosx=\frac{4}{5}\\cosx=\frac{-4}{5}\end{cases}}\)

- \(cosx=\frac{4}{5}\): 

\(sinx=cosxtanx=\frac{4}{5}.\frac{3}{4}=\frac{3}{5}\)

\(cotx=\frac{1}{tanx}=\frac{1}{\frac{3}{4}}=\frac{4}{3}\).

- \(cosx=\frac{-4}{5}\): 

\(sinx=cosxtanx=\frac{-4}{5}.\frac{3}{4}=\frac{-3}{5}\)

\(cotx=\frac{1}{tanx}=\frac{1}{\frac{3}{4}}=\frac{4}{3}\).

b)  \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-sin^2x=1-\frac{49}{625}=\frac{576}{625}\)

\(\Leftrightarrow\orbr{\begin{cases}cosx=\frac{24}{25}\\cosx=-\frac{24}{25}\end{cases}}\)

- \(cosx=\frac{24}{25}\): 

\(tanx=\frac{sinx}{cosx}=\frac{\frac{7}{25}}{\frac{24}{25}}=\frac{7}{24}\)

\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{\frac{7}{24}}=\frac{24}{7}\)

- \(cosx=\frac{-24}{25}\): 

\(tanx=\frac{sinx}{cosx}=\frac{\frac{7}{25}}{\frac{-24}{25}}=-\frac{7}{24}\)

\(tanx.cotx=1\Rightarrow cotx=\frac{1}{tanx}=\frac{1}{-\frac{7}{24}}=\frac{-24}{7}\)

a, Ta có: \(sin^2\alpha+cos^2\alpha=1\Leftrightarrow\left(\dfrac{3}{5}\right)^2+cos^2\alpha=1\Leftrightarrow cos\alpha=\pm\dfrac{4}{5}\)

Vậy đẳng thức có thể đồng thời xảy ra.

b, Ta có: \(1+cot^2\alpha=\dfrac{1}{sin^2\alpha}\Rightarrow1+cot^2\alpha=\dfrac{1}{\left(\dfrac{1}{3}\right)^2}\Rightarrow cot\alpha=\pm2\sqrt{2}\)

Hai đẳng thức không thể đồng thời xảy ra.

c, Ta có: \(tan\alpha\cdot cot\alpha=1\Rightarrow3\cdot cot\alpha=1\Rightarrow cot\alpha=\dfrac{1}{3}\)

Đẳng thức có thể đồng thời xảy ra.