Prove $\tanh x=\frac{\sinh(2x)}{1+\cosh(2x)}$

Question

Question

Prove $\tanh x=\frac{\sinh(2x)}{1+\cosh(2x)}$

My work:

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

Simplifies down to $$\frac{e^{2x}-1}{e^{2x}+1}$$

But then I do not know what to do from there

Answer 1

Dividing numerator and denominator by $e^x$ gives $$ \frac{e^{2x}-1}{e^{2x}+1} \frac{e^{-x}}{e^{-x}} = \frac{e^x-e^{-x}}{e^{x}+e^{-x}} = \frac{e^x-e^{-x}}{2}\frac{2}{e^{x}+e^{-x}} = \frac{\sinh{x}}{\cosh{x}}=\tanh{x}. $$