Question

Verify the formulas in Problems 17 to 24.

${\mathbf{tanh}}^{\mathbf{-}\mathbf{1}}\left(z\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{In}\left(\frac{1+z}{1-z}\right)$

Short answer

The formula ${\tanh }^{-1}\left(\mathrm{z}\right)=\frac{1}{2}\mathrm{In}\left(\frac{1+\mathrm{z}}{1-\mathrm{z}}\right)$is verified.

Step-by-step solution

Given Information

$\mathrm{Given}\mathrm{formula}\mathrm{is}{\tanh }^{-1}\left(\mathrm{z}\right)=\frac{1}{2}\mathrm{In}\left(\frac{1+\mathrm{z}}{1-\mathrm{z}}\right).$

Definition of Trigonometric equation.

A trigonometric equation is one that has one or more trigonometric ratios with unknown angles.

 Use exponential form to expand the equation

Given the function ${\tanh }^{-1}\left(\mathrm{z}\right)=\frac{1}{2}\mathrm{In}\left(\frac{1+\mathrm{z}}{1-\mathrm{z}}\right)$ .

Write the exponential form of the tanh(w)=z .

$$\begin{gathered} \tan h\left(w\right)=\frac{\sin h\left(w\right)}{\cos h\left(w\right)} \\ z=\frac{e^{\left(w\right)}+e^{\left(-w\right)}}{\left[e^{\left(w\right)}-e^{\left(w\right)}\right]}.......\left(1\right) \end{gathered}$$

Simplify equation (1).

$$\begin{gathered} {\mathrm{e}}^{\left(\mathrm{w}\right)}-{\mathrm{e}}^{\left(-\mathrm{w}\right)}={\mathrm{ze}}^{\left(\mathrm{w}\right)}+{\mathrm{ze}}^{\left(-\mathrm{w}\right)} \\ {\mathrm{e}}^{\left(2\mathrm{w}\right)}\left[1-\mathrm{z}\right]=\left[1+\mathrm{z}\right] \\ 2\mathrm{w}=\mathrm{In}\left(\frac{1+\mathrm{z}}{1-\mathrm{z}}\right) \\ \mathrm{w}=\frac{1}{2}\mathrm{In}\left(\frac{1+\mathrm{z}}{1-\mathrm{z}}\right) \end{gathered}$$

Replace value of w by w=$\tan h^{-1}\left(z\right)$ .

$\tan h^{-1}\left(z\right)=\frac{1}{2}In\left(\frac{1+z}{1-z}\right)$

Hence the formula is verified