Prove (a) cosh2(x) − sinh2(x) = 1 and (b) 1 − tanh 2(x) = sech 2(x). SOLUTION (a) cosh2(x) − sinh2(x) = ex + e−x 2 2 − 2 = e2x + 2 + e−2x 4 − = 4 = . (b) We start with the identity proved in part (a): cosh2(x) − sinh2(x) = 1. If we divide both sides by cosh2(x), we get 1 − sinh2(x) cosh2(x) = 1 or 1 − tanh 2(x) = .

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]