Răspuns: da, se verifica.
Explicație pas cu pas: se observa ca sunt egale.
[tex]\tan^{2}x+\cot^{2}x = \frac{\sin^{2}x}{\cos^{2}x}+\frac{cos^{2}x}{\sin^{2}x} = \frac{1}{\cos^2{x}}-1+\frac{1}{\sin^2{x}}-1 = \frac{1}{\sin^{2}x\cos^{2}x}-2=\frac{4-2\sin^{2}2x}{\sin^{2}2x};\\2\cdot\frac{3+\cos4x}{1-\cos4x}=2\cdot\frac{3+\cos^{2}2x-\sin^{2}2x}{1-\cos^{2}2x+\sin^{2}2x}=2\cdot\frac{3+1-\sin^{2}2x-\sin^{2}2x}{2\sin^{2}2x}=\\=\frac{4-2\sin^{2}2x}{\sin^{2}2x}[/tex]
