Question

Verify the formulas in Problems 17 to 24.

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

Short answer

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

Step-by-step solution

Given Information

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

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

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).

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

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

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

Hence the formula is verified.