Show that $cos h x + sin h x = e^{x}$ and simplify $cos h x - sin h x$
By multiplying the expressions for $(cos h x + sin h x)$ and $(cos h x - sin h x)$ together, show that ${cos}^{2} h x - {sin}^{2} h x = 1$

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Solution

We know that;
$c o s h x = \frac{e^{x} + e^{- x}}{2}$

$s i n h x = \frac{e^{x} - e^{- x}}{2}$
Thus;
$c o s h x + s i n h x = \frac{e^{x} + e^{- x}}{2} + \frac{e^{x} - e^{- x}}{2} = e^{x}$

$c o s h x - s i n h x = \frac{e^{x} + e^{- x}}{2} - \frac{e^{x} - e^{- x}}{2} = e^{- x}$
Thus;
$(c o s h x + s i n h x) \times (c o s h x - s i n h x) = e^{x} \times e^{- x}$

$\to c o s^{2} h x - s i n^{2} h x = 1$

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