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  "body": "\n## Proof\n\nWe expand both squares with the definitions.\n\n1. Let $x$ be a real number, and write $u = e^x$ and $v = e^{-x}$. Here $e^t = \\exp(t)$ is the [[exponential function|0RMVS7RXZX]].\n2. By the [[exponential of a sum|0E8HNVBCSP]], $u v = e^{x + (-x)} = e^0$.\n3. In the series defining $\\exp$, all terms with $n \\ge 1$ vanish at $0$. So $e^0 = 1$, and $u v = 1$ by step 2.\n4. By the definitions of the [[hyperbolic cosine|0EQFWKJ37J]] and the [[hyperbolic sine|0GWD21EWRW]], $\\cosh x = \\frac{u + v}{2}$ and $\\sinh x = \\frac{u - v}{2}$.\n5. Expanding the squares gives $(u + v)^2 = u^2 + 2uv + v^2$ and $(u - v)^2 = u^2 - 2uv + v^2$.\n6. By steps 4 and 5,\n$$\\cosh^2 x - \\sinh^2 x = \\frac{(u + v)^2 - (u - v)^2}{4} = \\frac{4 u v}{4} = u v.$$\n7. By steps 3 and 6, $\\cosh^2 x - \\sinh^2 x = 1$.\n\n## Idea\n\nBoth functions are built from $e^x$ and $e^{-x}$. In the difference of the squares, the $e^{2x}$ and $e^{-2x}$ terms cancel. Only the cross term $e^x e^{-x} = 1$ survives.\n",
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