Da expressão do enunciado: \begin{matrix} {{\dfrac{1}{(\cos^2{x}-\sin^2{x})^2}}} - {{ \left(\dfrac{2\tan{x}}{1-\tan^2{x}}\right)^2}} &=& {{\dfrac{1}{\cos^2{2x}}}} - \tan^2{2x} &=& {{\dfrac{1 - \sin^2{2x} } {\cos^2{2x}}}} &=&1 \end{matrix} \begin{matrix}Letra \ (C) \end{matrix}

Desenvolvendo a igualdade temos que $\dfrac{1}{(\cos^2x - \sin^2x)^2} - \dfrac{4\tan^2x}{(1 - \tan^2x)^2} = \dfrac{1}{(\cos2x)^2} - \dfrac{4\tan^2x}{\left(\dfrac{\cos^2x - \sin^2x}{\cos^2x}\right)^2}$ $=\dfrac{1}{\cos^2(2x)} - \dfrac{4\tan^2x}{\left(\dfrac{\cos2x}{\cos^2x}\right)^2} =\dfrac{1}{\cos^2(2x)} - \dfrac{4\tan^2x}{\cos^2(2x)} \cdot \cos^4x $ $= \dfrac{1}{\cos^2(2x)} - \dfrac{4\left(\dfrac{\sin^2x}{\cos^2x} \right) \cdot \cos^4x}{\cos^2(2x)} = \dfrac{1}{\cos^2(2x)} - \dfrac{4 \sin^2x \cdot \cos^2x}{\cos^2(2x)} $ $\dfrac{1}{\cos^2(2x)} - \dfrac{(2 \sin x \cdot \cos x)^2}{\cos^2(2x)}= \dfrac{1}{\cos^2(2x)} - \dfrac{\sin^2(2x)}{\cos^2(2x)} $ $=\dfrac{1 - \sin^2(2x)}{\cos^2(2x)} = \dfrac{\cos^2(2x)}{\cos^2(2x)} $ $= \boxed{\dfrac{1}{(\cos^2x - \sin^2x)^2} - \dfrac{4\tan^2x}{(1 - \tan^2x)^2} = 1}$ $\textbf{Resposta : Alternativa C}$