Can the following trigonometric equation be transformed into the other? Can $$16\sec^2⁡(x)\tan^4⁡(x)+88\sec^4⁡(x)\tan^2⁡(x)+16\sec^6⁡(x)$$ be proven equal to
$$24\sec^6(x)-8\sec^4(x)+96\sec^4(x)\tan^2(x)-16\sec^2(x)\tan^2(x)$$
I have made about six attempts, but I keep getting stuck. I thought I'd ask, maybe someone else can figure it out before me or verify that we cannot transform from one equation to another.
The reason why this is important to me is because I met a question asking for the differentiation of $\tan^2(x)$ 4 times. Both expressions above are different ways to express the SAME first order derivative, so they are indeed equal. However, the second expression has been derived by replacing $\tan^2(x)$ by $\sec^2(x)-1$, and then carrying on with differentiating to get the third and fourth derivatives. However, I didn't make such a substitution, hence I ended up with the first derivative. So, I'm trying to figure out a strategy to get the derivative right in the exam. It starts with knowing whether one of these expressions can somehow be converted into the other.

Sometimes the easiest thing to do is convert everything into sines and cosines.
\begin{array}{l}
   16 \sec^2⁡(x) \tan^4⁡(x) + 88 \sec^4⁡(x) \tan^2⁡(x) + 16 \sec^6⁡(x) \\
   =\dfrac{16\sin^4(x)}{\cos^6(x)}+\dfrac{88\sin^2(x)}{\cos^6(x)}
       +\dfrac{16}{\cos^6(x)} \\
   =\dfrac{16\sin^4(x) + 88 \sin^2(x) + 16}{\cos^6(x)}
\end{array}
\begin{array}{l}
   24\sec^6(x)-8\sec^4(x)+96\sec^4(x)\tan^2(x)-16\sec^2(x)\tan^2(x) \\
   =\dfrac{24}{\cos^6(x)}-\dfrac{8}{\cos^4(x)}+\dfrac{96\sin^2(x)}{\cos^6(x)}
       -\dfrac{16 \sin^2(x)}{\cos^4(x)} \\
   =\dfrac{24-8\cos^2(x)+96\sin^2(x)-16\sin^2(x)\cos^2(x)}{\cos^6(x)} \\
   =\dfrac{24-(8-8\sin^2(x))+96\sin^2(x)-(16\sin^2(x) - 16\sin^4(x))}{\cos^6(x)} \\
   =\dfrac{16\sin^4(x) + 88\sin^2(x) + 16}{\cos^6(x)}
\end{array}
