\(1+cot^2a=\dfrac{1}{sin^2a}\Rightarrow sin^2a=\dfrac{1}{1+cot^2a}=\dfrac{1}{1+4}=\dfrac{1}{5}\Rightarrow sina=\pm\dfrac{\sqrt{5}}{5}\)
\(TH1:sina=\dfrac{\sqrt{5}}{5}\)
\(B=sina+\dfrac{2cosa}{sin^3a-cos^3a}=sina+\dfrac{2cosa}{\left(sina-cosa\right)\left(sin^2a+cos^2a-sinacosa\right)}=sina+\dfrac{2cosa}{\left(sina-cosa\right)\left(1-sinacosa\right)}\)
\(cos^2a=1-sin^2a=1-\dfrac{1}{5}=\dfrac{4}{5}\Rightarrow cosa=\dfrac{2\sqrt[]{5}}{5}\left(cota=2>0\right)\)
\(B=sina+\dfrac{2cosa}{\left(sina-cosa\right)\left(1-sinacosa\right)}=\dfrac{\sqrt{5}}{5}+\dfrac{2.\dfrac{\sqrt{5}}{5}}{\left(\dfrac{\sqrt{5}}{5}-\dfrac{2\sqrt{5}}{5}\right)\left(1-\dfrac{\sqrt{5}}{5}.\dfrac{2\sqrt{5}}{5}\right)}\)
\(B=\dfrac{\sqrt{5}}{5}+\dfrac{2.\dfrac{\sqrt{5}}{5}}{\left(-\dfrac{\sqrt{5}}{5}\right)\left(1-\dfrac{2}{5}\right)}=\dfrac{\sqrt{5}}{5}+\dfrac{10}{3}=\dfrac{50+3\sqrt{5}}{15}\)
\(TH2:sina=-\dfrac{\sqrt{5}}{5}\Rightarrow cosa=-\dfrac{2\sqrt[]{5}}{5}\left(cota=2>0\right)\)
\(B=sina+\dfrac{2cosa}{\left(sina-cosa\right)\left(1-sinacosa\right)}=-\dfrac{\sqrt{5}}{5}+\dfrac{2.\left(-\dfrac{\sqrt{5}}{5}\right)}{\left(-\dfrac{\sqrt{5}}{5}+\dfrac{2\sqrt{5}}{5}\right)\left(1-\dfrac{-\sqrt{5}}{5}.\dfrac{-2\sqrt{5}}{5}\right)}\)
\(B=-\dfrac{\sqrt{5}}{5}+\dfrac{-2.\dfrac{\sqrt{5}}{5}}{\left(\dfrac{\sqrt{5}}{5}\right)\left(1-\dfrac{2}{5}\right)}=-\dfrac{\sqrt{5}}{5}-\dfrac{10}{3}=-\dfrac{\left(50+3\sqrt{5}\right)}{15}\)