Question

Please help me answer that question

Answer 1

By definitions of the hyperbolic functions, we have  proven that $cosh^{2}\theta-sinh^{2}\theta = 1$

Hyperbolic Functions & Proof of Identities

From the question, we are to prove that

ch²θ - sh²θ = 1

That is,

$cosh^{2}\theta-sinh^{2}\theta = 1$

By definition

$cosh\theta =\frac{e^{\theta}+ e^{-\theta}}{2}$

and

$sinh\theta =\frac{e^{\theta}- e^{-\theta}}{2}$

Thus,

$cosh^{2}\theta-sinh^{2}\theta$ becomes

$(\frac{e^{\theta}+ e^{-\theta}}{2})^{2} -(\frac{e^{\theta}- e^{-\theta}}{2})^{2}$

$\frac{(e^{\theta}+ e^{-\theta})^{2}}{2^{2}} -\frac{(e^{\theta}- e^{-\theta})^{2}}{2^{2}}$

$\frac{(e^{\theta}+ e^{-\theta})(e^{\theta}+ e^{-\theta})}{4} -\frac{(e^{\theta}- e^{-\theta})(e^{\theta}- e^{-\theta})}{4}$

$\frac{e^{2\theta}+ e^{0}+e^{0}+ e^{-2\theta}}{4} -\frac{(e^{2\theta}- e^{0}-e^{0}+e^{-2\theta})}{4}$

$\frac{e^{2\theta}+ 1+1+ e^{-2\theta}}{4} -\frac{(e^{2\theta}- 1-1+e^{-2\theta})}{4}$

$\frac{e^{2\theta}+2+ e^{-2\theta}}{4}-\frac{(e^{2\theta}- 2+e^{-2\theta})}{4}$

$\frac{e^{2\theta}+2+ e^{-2\theta}-(e^{2\theta}- 2+e^{-2\theta})}{4}$

$\frac{e^{2\theta}+2+ e^{-2\theta}-e^{2\theta}+2-e^{-2\theta}}{4}$

Collect like terms

$\frac{e^{2\theta}-e^{2\theta}+2+2+ e^{-2\theta}-e^{-2\theta}}{4}$

= $\frac{4}{4}$

= 1

Hence, By definitions of the hyperbolic functions, we have  proven that $cosh^{2}\theta-sinh^{2}\theta = 1$

Learn more Proving identities here: https://brainly.com/question/2561079

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