Show that $\frac{1-2^{-x}}{1+2^{-x}}$ is equivalent to $\frac{2^x-1}{2^x+1}$

Question

Show that $$\frac{1-2^{-x}}{1+2^{-x}}$$ is equivalent to $$\frac{2^x-1}{2^x+1}$$

I tried: $$\frac{1-2^{-x}}{1+2^{-x}} = \frac{1-\frac{1}{2^x}}{1+\frac{1}{2^x}}= \frac{\frac{2^x-1}{2^x}}{\frac{2^x+1}{2^x}}=\frac{2^{2x}-2^x}{2^{2x}+2^x}=???$$

What do I do next?

Answer 1

$$\frac{1-2^{-x}}{1+2^{-x}} = \frac{1-\frac{1}{2^x}}{1+\frac{1}{2^x}}= \frac{\frac{2^x-1}{2^x}}{\frac{2^x+1}{2^x}}=\frac{2^{x}-1}{2^{x}+1}$$

Answer 2

$$\frac{1-2^{-x}}{1+2^{-x}}=\frac{2^x}{2^x}\cdot\frac{1-2^{-x}}{1+2^{-x}}=\frac{2^x-2^{x-x}}{2^x+2^{x-x}}=\frac{2^x-1}{2^x+1}$$