91Ó°ÊÓ

Given equation is,$t\mathrm{an}z=\tan (x+iy)=\frac{\tan x+i\tan hy}{1-i\tan x\tan hy}$ .

A hyperbolic function is a representation of the relationship between a point's distances from the origin to the coordinate axes as a function of an angle.

Given the equation is,$\tan z=\tan (x+iy)=\frac{\tan x+i\tan hy}{1-i\tan x\tan hy}$ .

Write,$\tan z=\tan (x+yi)$.

Write,$\tan (x+yi)=\frac{\sin (x+yi)}{\cos (x+yi)}$. ….(1)

Use the following results.

localid="1658900147020" $$\begin{gathered} \sin (x+yi)=\sin (x)\cos h(y)+i\cos (x)\sin h(y) \\ \cos (x+yi)=\cos (x)\cos h(y)-i\sin (x)\sin h(y) \end{gathered}$$

Substitute the results in equation (1).

$$\begin{gathered} \tan (\mathrm{x}+\mathrm{yi})=\frac{\sin (\mathrm{x})\cosh (\mathrm{y})+\mathrm{i}\cos (\mathrm{x})\sinh (\mathrm{y})}{\cos (\mathrm{x})\cosh (\mathrm{y})-\mathrm{i}\sin (\mathrm{x})\sinh (\mathrm{y})}×\frac{1/\cosh \left(\mathrm{y}\right)}{1/\cosh \left(\mathrm{y}\right)} \\ =\frac{\sin (\mathrm{x})+\mathrm{i}\cos (\mathrm{x})\tanh \left(\mathrm{y}\right)}{\cos (\mathrm{x})\mathrm{i}\sin (\mathrm{x})\tanh \left(\mathrm{y}\right)}×\frac{1/\cos \left(\mathrm{x}\right)}{1/\cos \left(\mathrm{x}\right)} \\ =\frac{\tan (\mathrm{x})+\mathrm{itanh}\left(\mathrm{y}\right)}{1-\mathrm{i}\tan \left(\mathrm{x}\right)\tanh \left(\mathrm{y}\right)} \end{gathered}$$

Therefore, Left Hand Side(LHS) is equal to Right Hand Side(RHS).

Hence, the equation is verified.