\(B=\dfrac{2sin2a-sin4a}{2sin2a+sin4a}=\dfrac{2sin2a-2sin2a.cos2a}{2sin2a+2sin2a.cos2a}\)
\(=\dfrac{2sin2a\left(1-cos2a\right)}{2sin2a\left(1+cos2a\right)}=\dfrac{1-cos2a}{1+cos2a}=\dfrac{1-\left(1-2sin^2a\right)}{1+2cos^2a-1}\)
\(=\dfrac{2sin^2a}{2cos^2a}=tan^2a\)