Question

[Lê Nguyễn Vân An]

giúp tớ với 

Answer 1

[YangSu]

\(a,cos\alpha=\dfrac{5}{13}\)

\(sin\alpha=\sqrt{1-cos^2\alpha}=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)

\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+tan^2\alpha=\dfrac{1}{\left(\dfrac{5}{13}\right)^2}\Leftrightarrow tan^2\alpha=\dfrac{144}{25}\Leftrightarrow tan\alpha=\dfrac{12}{5}\)

\(cot\alpha=\dfrac{1}{tan\alpha}=1:\dfrac{12}{5}=\dfrac{5}{12}\)

\(b,sin\alpha=\dfrac{7}{12}\)

\(cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{1-\left(\dfrac{7}{12}\right)^2}=\dfrac{\sqrt{95}}{12}\)

\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+tan^2\alpha=\dfrac{1}{\left(\dfrac{\sqrt{95}}{12}\right)^2}\Leftrightarrow tan\alpha=\dfrac{49}{95}\)

\(cot\alpha=1:\dfrac{49}{95}=\dfrac{95}{49}\)

\(c,tan\alpha=\dfrac{15}{4}\)

\(cot\alpha=1:\dfrac{15}{4}=\dfrac{4}{15}\)

\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+\left(\dfrac{15}{4}\right)^2=\dfrac{1}{cos^2\alpha}\Leftrightarrow cos\alpha=\sqrt{\dfrac{16}{241}}\)

\(sin\alpha=\sqrt{1-cos^2\alpha}=\sqrt{1-\left(\sqrt{\dfrac{16}{241}}\right)^2}\approx0,97\)

\(d,cot\alpha=-\dfrac{1}{\sqrt{3}}\\ tan\alpha=1:\left(-\dfrac{1}{\sqrt{3}}\right)=-\sqrt{3}\)

\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+\left(-\sqrt{3}\right)^2=\dfrac{1}{cos^2\alpha}\Leftrightarrow cos\alpha=\dfrac{1}{2}\)

\(sin\alpha=\sqrt{1-\left(\dfrac{1}{2}\right)^2}=\dfrac{\sqrt{3}}{2}\)