91Ó°ÊÓ

Given equation is${\cosh }^2\mathrm{z}-{\sinh }^2\mathrm{z}=1$

The term "Hyperbolic Function" refers to the relationship between a point on a hyperbola's distance from its origin and its coordinate axes, expressed as a function of an angle.

Write the exponential form of the given equation.

$\sin h\left(z\right)=\frac{e^{\left(z\right)}-e^{\left(-z\right)}}{2}$ …(1)

$\cos h\left(z\right)=\frac{e^{\left(z\right)}+e^{\left(-z\right)}}{2}$ …(2)

Square both the exponential form i.e. equation (1) and (2).

${\sinh }^2\left(\mathrm{z}\right)=\frac{1}{4}{\left[\exp \left(\mathrm{z}\right)-\exp \left(-\mathrm{z}\right)\right]}^2$

${\sinh }^2\left(\mathrm{z}\right)=\frac{1}{4}\left[\exp \left(2\mathrm{z}\right)+\exp \left(-2\mathrm{z}\right)-2\right]$ …(3)

$\cos h^2\left(\mathrm{z}\right)=\frac{1}{4}{\left[\exp \left(\mathrm{z}\right)+\exp \left(-\mathrm{z}\right)\right]}^2$

$\cos h^2\left(\mathrm{z}\right)=\frac{1}{4}\left[\exp \left(2\mathrm{z}\right)+\exp \left(-2\mathrm{z}\right)+2\right]$ …(4)

Subtract the square terms of exponential form i.e. equation (3) and (4).

$$\begin{gathered} \cos h^2\left(\mathrm{z}\right)-\sin h^2\left(z\right)=\frac{1}{4}\left[\exp \left(2\mathrm{z}\right)+\exp \left(-2\mathrm{z}\right)+2\right]-\frac{1}{4}\left[\exp \left(2\mathrm{z}\right)+\exp \left(-2\mathrm{z}\right)+2\right] \\ =\frac{1}{4}\left[4\right] \\ =1 \end{gathered}$$

Hence the equation is verified.