<s>[INST] using trigonometric identities For proving $$\frac {16}{\cos (4x)+7} =\frac{1}{\sin^4x +\cos^2x} +\frac{1}{\sin^2x +\cos^4x} $$
I tried to use that:
\begin{align}
\sin^4 x +\cos^4 x&=\sin^4 x +2\sin^2x\cos^2 x+\cos^4 x - 2\sin^2x\cos^2 x\\
&=(\sin^2x+\cos^2 x)^2-2\sin^2x\cos^2 x\\
&=1^2-\frac{1}{2}(2\sin x\cos x)^2\\
&=1-\frac{1}{2}\sin^2 (2x)\\
&=1-\frac{1}{2}\left(\frac{1-\cos 4x}{2}\right)\\
&=\frac{3}{4}+\frac{1}{4}\cos 4x
\end{align}
but i can't try more
 [/INST] Note that
$$\begin{align}\sin^4x +\cos^2x &=
\sin^2x(1-\cos^2x) +\cos^2x\\
&=\sin^2x+\cos^2x(1-\sin^2x)=\sin^2x +\cos^4x. 
\end{align}$$
Hence, according to your work, 
$$\begin{align}
2(\sin^4x +\cos^2x)&=(\sin^4x +\cos^2x) +(\sin^2x +\cos^4x)\\
&=\sin^4x +\cos^4x+1=\frac{7+\cos 4x}{4}.\end{align}$$
Can you take it from here?
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