Answer:
See explanation.
Step-by-step explanation:
The hyperbolic sine and cosine functions are defined as follows:
[tex] \sinh(x) = \frac{ {e}^{x} - {e}^{ - x} }{2} [/tex]
[tex]\cosh(x) = \frac{ {e}^{x} + {e}^{ - x} }{2} [/tex]
We want to show that:
[tex]\cosh^{2} (x) - \sinh^{2} (x) = 1[/tex]
We use the definition by substituting the expressions into the left hand side and simplify to obtain the RHS.
[tex] \cosh^{2} (x) - \sinh^{2} (x) = {( \frac{ {e}^{x} + {e}^{ - x} }{2} )}^{2} + {( \frac{ {e}^{x} - {e}^{ - x} }{2} )}^{2} [/tex]
[tex]\cosh^{2} (x) - \sinh^{2} (x) = \frac{ {e}^{2x} +2 {e}^{x} {e}^{ - x} + {e}^{ - 2x} }{4} - \frac{ {e}^{2x} - 2 {e}^{x} {e}^{ - x} + {e}^{ - 2x} }{4} [/tex]
[tex]\cosh^{2} (x) - \sinh^{2} (x) = \frac{ {e}^{2x} +2 + {e}^{ - 2x} }{4} - \frac{ {e}^{2x} - 2 + {e}^{ - 2x} }{4} [/tex]
[tex]\cosh^{2} (x) - \sinh^{2} (x) = \frac{ {e}^{2x} + {e}^{ - 2x} - {e}^{2x} + {e}^{ - 2x }+ 2 +2 }{4} [/tex]
[tex]\cosh^{2} (x) - \sinh^{2} (x) = \frac{ 4 }{4} [/tex]
[tex]\cosh^{2} (x) - \sinh^{2} (x)=1[/tex]
b)
If we start with the identity in a) and we divide both sides by cosh²x we get:
[tex] \frac{\cosh^{2} (x) }{\cosh^{2} (x) } -\frac{\sinh^{2} (x) }{\cosh^{2} (x) } =\frac{1}{\cosh^{2} (x) } [/tex]
This simplifies to:
[tex]1 - \tanh ^{2} (x) = \sec \: h ^{2} (x) [/tex]
