<s>[INST] Evaluate $\lim_{h\to 0}\frac{1}{h^2}\begin{vmatrix}\tan x&\tan(x+h)&\tan(x+2h)\\\tan(x+2h)&\tan x&\tan(x+h)\\\tan(x+h)&\tan(x+2h)&\tan x\end{vmatrix}$ Evaluate
$$
\lim_{h\to 0}\frac{\Delta}{h^2}=\lim_{h\to 0}\frac{1}{h^2}\begin{vmatrix}
\tan x&\tan(x+h)&\tan(x+2h)\\
\tan(x+2h)&\tan x&\tan(x+h)\\
\tan(x+h)&\tan(x+2h)&\tan x
\end{vmatrix}
$$
Attempt
$$
\lim_{h\to 0}\frac{\Delta}{h^2}=\begin{vmatrix}
\lim_{h\to 0}\tan x&\lim_{h\to 0}\dfrac{\tan(x+h)-\tan x}{h}&\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}\\
\lim_{h\to 0}\tan(x+2h)&\lim_{h\to 0}\dfrac{\tan x-\tan(x+2h)}{h}&\lim_{h\to 0}\dfrac{\tan(x+h)-\tan(x+2h)}{h}\\
\lim_{h\to 0}\tan(x+h)&\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}&\lim_{h\to 0}\dfrac{\tan x-\tan(x+2h)}{h}
\end{vmatrix}\\
=\begin{vmatrix}
\lim_{h\to 0}\tan x&\lim_{h\to 0}\dfrac{\tan(x+h)-\tan x}{h}&\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}\\
\lim_{h\to 0}\tan(x+2h)&-2.\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan x}{h}&-1.\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}\\
\lim_{h\to 0}\tan(x+h)&\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}&-2.\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan x}{2h}
\end{vmatrix}\\
$$
$$
\lim_{h\to 0}\dfrac{\tan(x+h)-\tan x}{h}=\frac{d}{dx}\tan x=\sec^2x\\
\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan x}{2h}=\frac{d}{dx}\tan x=\sec^2x\\
\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}=\frac{d}{dx}\tan(x+h)=\sec^2(x+h)
$$
But my reference gives the solution $9\tan x.\sec^4x$, I think by taking $\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}=\sec^2x$. Will that make a difference ?
It might be silly but could anyone clarify this confusion in my attempt ?
 [/INST] For
$$M=\left(
\begin{array}{ccc}
 a & b & c \\
 c & a & b \\
 b & c & a
\end{array}
\right)\implies |M|=a^3+b^3+c^3-3 a b c$$ So, replacing
$$\Delta=\tan ^3(x)+\tan ^3(x+h)+\tan ^3(x+2 h)-3 \tan (x) \tan (x+h) \tan (x+2 h)$$ Now, composing Taylor series around $h=0$ gives
$$\Delta=9  \left(\tan ^5(x)+2 \tan ^3(x)+\tan (x)\right)h^2+O\left(h^3\right)=9\tan (x) \sec ^4(x)\,h^2+O\left(h^3\right)$$
</s>