\(a,cos\alpha=\dfrac{5}{13}\)
\(sin\alpha=\sqrt{1-cos^2\alpha}=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)
\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+tan^2\alpha=\dfrac{1}{\left(\dfrac{5}{13}\right)^2}\Leftrightarrow tan^2\alpha=\dfrac{144}{25}\Leftrightarrow tan\alpha=\dfrac{12}{5}\)
\(cot\alpha=\dfrac{1}{tan\alpha}=1:\dfrac{12}{5}=\dfrac{5}{12}\)
\(b,sin\alpha=\dfrac{7}{12}\)
\(cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{1-\left(\dfrac{7}{12}\right)^2}=\dfrac{\sqrt{95}}{12}\)
\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+tan^2\alpha=\dfrac{1}{\left(\dfrac{\sqrt{95}}{12}\right)^2}\Leftrightarrow tan\alpha=\dfrac{49}{95}\)
\(cot\alpha=1:\dfrac{49}{95}=\dfrac{95}{49}\)
\(c,tan\alpha=\dfrac{15}{4}\)
\(cot\alpha=1:\dfrac{15}{4}=\dfrac{4}{15}\)
\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+\left(\dfrac{15}{4}\right)^2=\dfrac{1}{cos^2\alpha}\Leftrightarrow cos\alpha=\sqrt{\dfrac{16}{241}}\)
\(sin\alpha=\sqrt{1-cos^2\alpha}=\sqrt{1-\left(\sqrt{\dfrac{16}{241}}\right)^2}\approx0,97\)
\(d,cot\alpha=-\dfrac{1}{\sqrt{3}}\\ tan\alpha=1:\left(-\dfrac{1}{\sqrt{3}}\right)=-\sqrt{3}\)
\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+\left(-\sqrt{3}\right)^2=\dfrac{1}{cos^2\alpha}\Leftrightarrow cos\alpha=\dfrac{1}{2}\)
\(sin\alpha=\sqrt{1-\left(\dfrac{1}{2}\right)^2}=\dfrac{\sqrt{3}}{2}\)