Questions mathématiques diverses

Question 1:
Simplifier
$\sqrt{\sin^4 x+4\cos^2 x}-\sqrt{\cos^4 x+4\sin^2 x}$

Réponse 1:
$$\begin{align*} &\sqrt{\sin^4 x+4\cos^2x}-\sqrt{\cos^4x+4\sin^2x}\\ &=\sqrt{\sin^4x+4(1-\sin^2x)}-\sqrt{\cos^4x+4(1-\cos^2x)}\\ &=\sqrt{\sin^4x-4\sin^2x+4}-\sqrt{\cos^4x-4\cos^2x+4}\\ &=\sqrt{(\sin^2x-2)^2}-\sqrt{(\cos^2x-2)^2}\\ &=\left|{\sin^2x-2}\right|-\left|{\cos^2x-2}\right|\\ &=(2-\sin^2x)-(2-\cos^2x)\\ &=\cos^2x-\sin^2x\\ &=\cos 2x \end{align*}$$