Simplifying with exponents

**DivideBy0** · July 11th 2007, 10:55 PM

Mr. Math Book says that

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

Can someone tell me how the LHS was simplified?? Thx

**red_dog** · July 11th 2007, 11:03 PM

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

**Soroban** · July 12th 2007, 03:39 AM

Hello, DivideBy0!

Another way . . .

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

Multiply by $\frac{e^x}{e^x}$

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

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