cot x=2>0
=>sin x và cosx cùng dấu
=>sinx*cosx>0
\(1+cot^2x=\dfrac{1}{sin^2x}=1+4=5\)
=>sin^2x=1/5
=>cos^2x=4/5
\(B=\dfrac{1}{5}-2\cdot sinx\cdot cosx-\dfrac{1}{5}\cdot\dfrac{4}{5}+\dfrac{1}{5}-3\)
\(=\dfrac{2}{5}-\dfrac{4}{25}-3-2\cdot\dfrac{1}{\sqrt{5}}\cdot\dfrac{2}{\sqrt{5}}\)
\(=\dfrac{10}{25}-\dfrac{4}{25}-\dfrac{75}{25}-2\cdot\dfrac{2}{5}=\dfrac{-69}{25}-\dfrac{4}{5}=\dfrac{-89}{25}\)
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