Solution:

(i) $1+\mathrm{i} \tan \alpha$
Given that $Z=1+\mathrm{i}$ tan $\alpha$
It is known to us that the polar form of a complex number $Z=$
In which,
$\begin{array}{l}
\left.|Z|=\text { modulus of complex number }=\sqrt{(} x^{2}+y^{2}\right) \\
\theta=\arg (z)=\text { argument of complex number }=\tan ^{-1}(|y| /|x|)
\end{array}$
It is also known that tan $\alpha$ is a periodic function with per
Therefore $\alpha$ is lying in the interval $[0, \pi / 2) \cup(\pi / 2, \Pi]$.
Consider case 1 :
$a \in[0, \pi / 2)$
Now,
$\begin{array}{l}
\left.|Z|=r=\sqrt{(} x^{2}+y^{2}\right) \\
=\sqrt{\left(1^{2}+\tan ^{2} a\right)} \\
{=} \sqrt{\left(\sec ^{2} \alpha\right)} \\
{=} &{|\sec \alpha| \text { since, }} \\
{\theta=\tan ^{-1}(|y| /|x|)} \\
{=\tan ^{-1}(\tan \alpha / 1)} \\
{=\tan ^{-1}(\tan \alpha)}
\end{array}$
$=\alpha$ asa, tan $a$ is positive in the interval $[6$ $\therefore$ The polar form is $Z=\sec \alpha(\cos \alpha+i \sin a)$
Consider case 2:
$\alpha \in(\pi / 2, \pi]$
Now,
$\begin{array}{l}
\left.|Z|=r=\sqrt{(} x^{2}+y^{2}\right) \\
=\sqrt{\left(1^{2}+\tan ^{2} a\right)} \\
{=} \sqrt{\left(\sec ^{2} a\right)} \\
{=} {\mid \sec a}|
\end{array}$

(ii) $\tan \alpha-i$
Given that $Z=\tan \alpha-i$
It is known to us that the polar form of a complex number $Z=x+i$ iy is given by $Z=|Z|(\cos \theta+i \sin \theta)$
In which,
$\begin{array}{l}
\left.|Z|=\text { modulus of complex number }=\sqrt{(} x^{2}+y^{2}\right) \\
\theta=\arg (z)=\text { argument of complex number }=\tan ^{-1}(|y| /|x|)
\end{array}$
It is also known that tan $\alpha$ is a periodic function with period $\pi .$
Therefore $\alpha$ is lying in the interval $[0, \pi / 2) \cup(\pi / 2, \pi]$.
Consider case 1:
$\alpha \in[0, \pi / 2)$
Now,
$|Z|=r=\sqrt{\left(x^{2}+y^{2}\right)}$
$=\sqrt{\left(\tan ^{2} \alpha+1^{2}\right)}$
$\left.=\sqrt{\left(\sec ^{2}\right.} \mathrm{a}\right)$
$=\mid$ sec $a \mid$ since, sec $a$ is positive in the interval $[0, \pi / 2)$
$=\sec \alpha$
$\theta=\tan ^{-1}(|y| /|x|)$
$=\tan ^{-1}(1 / \tan \alpha)$
$=\tan ^{-1}(\cot \alpha)$ since, $\cot \alpha$ is positive in the interval $[0, \pi / 2)$
$=\alpha-\pi / 2\left[\right.$ since, $\theta$ lies in $4^{\text {th }}$ quadrant]
$Z=\sec \alpha(\cos (\alpha-\pi / 2)+i \sin (\alpha-\pi / 2))$
$\therefore$ Polar form is $Z=\sec \alpha(\cos (\alpha-\pi / 2)+i \sin (\alpha-\pi / 2))$
Consider case $2:$
$a \in(\Pi / 2, \Pi]$
Now,
$\begin{array}{l}
\left.|Z|=r=\sqrt{(} x^{2}+y^{2}\right) \\
\left.=\sqrt{\left(\tan ^{2}\right.} a+1^{2}\right) \\
=\sqrt{\left(\sec ^{2} a\right)} \\
{=} {|\sec a|}
\end{array}$
$=-\sec \alpha$ since, sec $\alpha$ is negative in the interval $(\pi / 2, \pi]$
$\begin{array}{l}
\theta=\tan ^{-1}(|\mathrm{y}| /|\mathrm{x}|) \\
=\tan ^{-1}(1 / \tan \alpha) \\
=\tan ^{-1}(\cot \alpha)
\end{array}$
$=\pi / 2+a$ since, cot $a$ is negative in the interval $(\pi / 2, \pi]$
$\theta=\pi / 2+\alpha\left[\right.$ since, $\theta$ lies in $3^{\text {th }}$ quadrant]
$Z=-\sec \alpha(\cos (\pi / 2+\alpha)+i \sin (\pi / 2+a))$

As a result, Polar form is $Z=-\sec \alpha(\cos (\pi / 2+a)+i \sin (\pi / 2+a))$