We have to find:
$$\tan x\left(1-\sec \frac{x}{2}\right)\prod_{r=1}^{8} \left(1-\sec(2^{r-1}x)\right)$$
Write:
$$1-\sec(2^{r-1}x)=1-\frac{2}{z^{2^{r-1}}+z^{-2^{r-1}}}=1-\frac{2z^{2^{r-1}}}{1+z^{2^r}}=\frac{\left(1-z^{2^{r-1}}\right)^2}{1+z^{2^r}}$$
where $z=e^{ix}$.
$$\Rightarrow \tan x\left(1-\sec \frac{x}{2}\right)\prod_{r=1}^{8} \left(1-\sec(2^{r-1}x)\right)=\tan x\left(1-\sec \frac{x}{2}\right)\dfrac{\displaystyle \left(\prod_{r=1}^{8} \left(1-z^{2^{r-1}}\right)\right)^2}{\displaystyle \prod_{r=1}^{8} \left(1+z^{2^r}\right)}$$
$$\large \prod_{r=1}^{8} \left(1-z^{2^{r-1}}\right)=1-z+z^2-z^3+\cdots -z^{2^8-1}=\frac{1-z^{2^8}}{1+z}$$
$$\large \prod_{r=1}^{8} \left(1+z^{2^r}\right)=1+z^2+z^4+\cdots+z^{2(2^8-1)}=\frac{\left(1-z^{2^9}\right)}{1-z^2}=\frac{\left(1-z^{2^8}\right)\left(1+z^{2^8}\right)}{(1-z)(1+z)}$$
Hence,
$$\tan x\left(1-\sec \frac{x}{2}\right)\dfrac{\displaystyle \left(\prod_{r=1}^{8} \left(1-z^{2^{r-1}}\right)\right)^2}{\displaystyle \prod_{r=1}^{8} \left(1+z^{2^r}\right)}=\tan x\left(1-\sec \frac{x}{2}\right)\frac{1-z}{1+z}\frac{1-z^{2^8}}{1+z^{2^8}}$$
Substitute back $z=e^{ix}$ and simplify in the following way:
$$\frac{1-z}{1+z}=\frac{1-e^{ix}}{1+e^{ix}}=\frac{e^{-ix/2}-e^{ix/2}}{e^{-ix/2}+e^{ix/2}}=-i\tan\frac{x}{2}$$
Similarly,
$$\frac{1-z^{2^8}}{1+z^{2^8}}=-i\tan(2^7x)$$
$$\Rightarrow \tan x\left(1-\sec \frac{x}{2}\right)\prod_{r=1}^{8} \left(1-\sec(2^{r-1}x)\right)=-\tan x\left(1-\sec \frac{x}{2}\right)\tan\frac{x}{2}\tan(2^7x)$$
Use:
$$1-\sec\frac{x}{2}=-\frac{2\sin^2\frac{x}{4}}{\cos\frac{x}{2}}$$
to obtain:
$$2\frac{\sin^2\frac{x}{4}}{\cos\frac{x}{2}}\tan\frac{x}{2}\tan x\tan(2^7x)$$
Do I have to simplify further?
(Doh)
