14. Ta có \(sin^2\alpha+cos^2\alpha=1\Rightarrow\left(\dfrac{5}{13}\right)^2+cos^2\alpha=1\Rightarrow cos^2\alpha=1-\left(\dfrac{5}{13}\right)^2=\dfrac{144}{169}\Rightarrow cos\alpha=\sqrt{\dfrac{144}{169}}=\dfrac{12}{13}\)
\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{5}{13}:\dfrac{12}{13}=\dfrac{5}{12}\)
\(cot\alpha=\dfrac{1}{tan\alpha}=1:\dfrac{5}{12}=\dfrac{12}{5}\)
15. \(A=sin^210^o+sin^220^o+...sin^260^o+sin^280^o=sin^210^o+sin^220^o+sin^230^o+sin^240^o+cos^240^o+cos^230^o+cos^220^o+cos^210^o=\left(sin^210^o+cos^210^o\right)+\left(sin^220^o+cos^220^o\right)+\left(sin^230^o+cos^230^o\right)+\left(sin^240^o+cos^240^o\right)=1+1+1+1=4\)
\(B=cos^215^o-cos^225^o+cos^235^o-cos^245^o+cos^255^o-cos^265^o+cos^275^o=cos^215^o-sin^265^o+cos^235^o-cos^245^o+sin^235^o-cos^265^o+sin^215^o=\left(sin^215^o+cos^215^o\right)+\left(sin^235^o+cos^235^o\right)-\left(sin^265^o+cos^265^o\right)-cos^245^o=1+1-1-cos^245^o=1-\left(\dfrac{\sqrt{2}}{2}\right)^2=1-\dfrac{1}{2}=\dfrac{1}{2}\)\(C=tan20^o.tan40^o.tan50^o.tan60^o.tan70^o=tan20^o.tan40^o.tan50^o.cot40^o.cot20^o=\left(tan20^o.cot20^o\right)\left(tan40^o.cot40^o\right)tan50^o=1.1.tan50^o=tan50^o\simeq1,19\)
