# 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.

• WolframAlpha confirms that the two are equal Jun 6, 2017 at 13:43
• Also, it may be helpful to people if you state what your 6 attempts were, and why they didn't work, so that people don't repeat your attempts, or can point out where you went wrong Jun 6, 2017 at 13:45
• They are too tedious and worthless to write out because they are mostly flawed with arithmetic error. Also see the edit to my question. Moreover, here is my latest attempt: take the sec^2(x)tan^4(x) term and change the sec^2(x) to 1+tan^2(x) and change the the tan^4(x) to [1+tan^2(x)]^2. Then I expanded and compared with the second expression and what alterations it needs to undergo in order to match the first expression. Then I decided to ask a question here because the stack exchange series is LITERALLY (not figuratively) the most useful thing in the world right now. Jun 6, 2017 at 13:59

\begin{align} &16\sec^2⁡x\color{red}{\tan^4⁡x}+88\sec^4x\tan^2x+16\sec^6x\\ &16\sec^2⁡x\color{red}{(\sec^2x-1)^2}+88\sec^4x\tan^2x+16\sec^6x\\ &16\sec^2⁡x\color{red}{(\sec^4x-2\sec^2x+1)}+88\sec^4x\tan^2x+16\sec^6x\\ &16\sec^6⁡x-32\sec^4x+16\sec^2x+88\sec^4x\tan^2x+16\sec^6x\\ &\color{blue}{32\sec^6⁡x}+\color{green}{(-32\sec^4x)}+16\sec^2x+88\sec^4x\tan^2x\\ &\color{blue}{(24\sec^6⁡x + 8\sec^6⁡x)}+\color{green}{(-8\sec^4x -24\sec^4x) }+16\sec^2x+88\sec^4x\tan^2x\\ &24\sec^6⁡x + 8\sec^4⁡x(\color{red}{\sec^2x)}-8\sec^4x -24(\tan^2x+1)\sec^2x+16\sec^2x+88\sec^4x\tan^2x\\ &24\sec^6⁡x + 8\sec^4⁡x\color{red}{(\tan^2x+1)}-8\sec^4x -24\tan^2x\sec^2x-24\sec^2x+16\sec^2x+88\sec^4x\tan^2x\\ &24\sec^6⁡x + 8\sec^4⁡x\tan^2x+8\sec^4⁡x-8\sec^4x -24\tan^2\sec^2x-8\sec^2x+88\sec^4x\tan^2\\ &24\sec^6⁡x -8\sec^4x + 96\sec^4⁡x\tan^2x +(\color{green}{8\sec^4x}-24\tan^2\sec^2x-8\sec^2x)\\ &24\sec^6⁡x -8\sec^4x + 96\sec^4⁡x\tan^2x +(\color{green}{8(\tan^2x+1)\sec^2x} -24\tan^2\sec^2x-8\sec^2x)\\ &24\sec^6⁡x -8\sec^4x + 96\sec^4⁡x\tan^2x +\color{green}{8\tan^2x\sec^2x+8\sec^2x}-24\tan^2\sec^2x-8\sec^2x\\ &24\sec^6x-8\sec^4x+96\sec^4x\tan^2x-16\sec^2x\tan^2x \end{align}

• Well done man :) Jun 6, 2017 at 14:34

Let me edit it. So i write $\sec (x)=s,\tan (x)=t$ thus we have $16s^2t^4+88s^4t^2+16s^6$ so writing last term as then we write it as $16s^2t^2 (s^2-1)+88s^4t^2+8s^6+8s^4t^2+8s^4=16s^2t^4+96s^4t^2+8s^6+8s^4=16s^2t^2 (s^2-1)+96s^4t^2+8s^6+8s^4=16s^4t^2-16s^2t^2+96s^4t^2+8s^6+8s^4=16s^4 (s^2-1)+96s^4t^2-16s^2t^2+8s^6+8s^4=24s^6-8s^4+96s^4t^2-16s^2t^2$

• That is what I have been attempting. May you please try to write out a solution if you don't mind lending me some of your thinking time? Jun 6, 2017 at 13:45
• I have edited I just hurried a bit when I saw it first time so ignore what I said first. Jun 6, 2017 at 14:17

Sometimes the easiest thing to do is convert everything into sines and cosines.

$$\begin{array}{l} 16 \sec^2⁡(x) \tan^4⁡(x) + 88 \sec^4⁡(x) \tan^2⁡(x) + 16 \sec^6⁡(x) \\ =\dfrac{16\sin^4(x)}{\cos^6(x)}+\dfrac{88\sin^2(x)}{\cos^6(x)} +\dfrac{16}{\cos^6(x)} \\ =\dfrac{16\sin^4(x) + 88 \sin^2(x) + 16}{\cos^6(x)} \end{array}$$

$$\begin{array}{l} 24\sec^6(x)-8\sec^4(x)+96\sec^4(x)\tan^2(x)-16\sec^2(x)\tan^2(x) \\ =\dfrac{24}{\cos^6(x)}-\dfrac{8}{\cos^4(x)}+\dfrac{96\sin^2(x)}{\cos^6(x)} -\dfrac{16 \sin^2(x)}{\cos^4(x)} \\ =\dfrac{24-8\cos^2(x)+96\sin^2(x)-16\sin^2(x)\cos^2(x)}{\cos^6(x)} \\ =\dfrac{24-(8-8\sin^2(x))+96\sin^2(x)-(16\sin^2(x) - 16\sin^4(x))}{\cos^6(x)} \\ =\dfrac{16\sin^4(x) + 88\sin^2(x) + 16}{\cos^6(x)} \end{array}$$