Answer:
See below
Step-by-step explanation:
By definition:
[tex] \sinh x = \frac{ {e}^{x} + {e}^{ - x} }{2} \\ \cosh x = \frac{ {e}^{x} - {e}^{ - x} }{2} [/tex]
We want to show that:
[tex] { \sinh}^{2} x - { \cosh}^{2}x = 1 [/tex]
We substitute into the left side to get:
[tex]( \frac{ {e}^{x} + {e}^{ - x} }{2}) ^{2} - ( \frac{ {e}^{x} - {e}^{ - x} }{2}) ^{2} [/tex]
We apply exponential rule to get:
[tex] = \frac{ ({e}^{x} + {e}^{ - x})^{2} }{4} - \frac{( {e}^{x} - {e}^{ - x} ) ^{2} }{4}[/tex]
We expand:
[tex] = \frac{ {e}^{2x} + 2 {e}^{ - x} {e}^{x} + {e}^{ - 2x} }{4} - \frac{ {e}^{2x} - 2 {e}^{ - x} {e}^{x} + {e}^{ - 2x} }{4} [/tex]
[tex] = \frac{ {e}^{2x} + 2 + {e}^{ - 2x} }{4} - \frac{ {e}^{2x} - 2 + {e}^{ - 2x} }{4} [/tex]
[tex] = \frac{ {e}^{2x} + 2 + {e}^{ - 2x} - {e}^{2x} + 2 - {e}^{ - 2x}}{4} [/tex]
[tex] = \frac{4}{4} \\ = 1[/tex]