cotx=2
=>cosx=2*sin x
\(1+cot^2x=\dfrac{1}{sin^2x}\)
=>\(\dfrac{1}{sin^2x}=1+4=5\)
=>\(sin^2x=\dfrac{1}{5}\)
\(B=\dfrac{sin^2x-2\cdot sinx\cdot2\cdot sinx-1}{5\cdot4sin^2x+sin^2x-3}=\dfrac{-3sin^2x-1}{21sin^2x-3}\)
\(=\dfrac{-\dfrac{3}{5}-1}{\dfrac{21}{5}-3}=-\dfrac{8}{5}:\dfrac{6}{5}=-\dfrac{4}{3}\)
\(cotx=2\Rightarrow tanx=\dfrac{1}{2}\)
\(B=\dfrac{sin^2x-2sinx.cosx-1}{5cos^2x+sin^2x-3}\)
\(\Leftrightarrow B=\dfrac{tan^2x-2tanx-\dfrac{1}{cos^2x}}{5+tan^2x-\dfrac{3}{cos^2x}}\)
\(\Leftrightarrow B=\dfrac{tan^2x-2tanx-1-tan^2x}{5+tan^2x-3-3tan^2x}\)
\(\Leftrightarrow B=\dfrac{-2tanx-1}{2-2tan^2x}\)
\(\Leftrightarrow B=\dfrac{-2.\dfrac{1}{2}-1}{2-2.\dfrac{1}{4}}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\)
