\(\frac{sin^22x+4sin^2x-4}{sin^22x-4sin^2x}=\frac{4sin^2x.cos^2x-4\left(1-sin^2x\right)}{4sin^2x.cos^2x-4sin^2x}=\frac{4sin^2x.cos^2x-4cos^2x}{4sin^2x.cos^2x-4sin^2x}\)
\(=\frac{cos^2x\left(sin^2x-1\right)}{sin^2x\left(cos^2x-1\right)}=\frac{cos^2x.\left(-cos^2x\right)}{sin^2x\left(-sin^2x\right)}=\frac{cos^4x}{sin^4x}=cot^4x\)