Answer
$\tan{165^o}=\tan{(135^o+30^o)}=-2+\sqrt3$
Work Step by Step
Note that $165^o=135^o+30^o$.
Hence, the given expression is equivalent to $\tan(135^o+30^o)$.
RECALL:
$\tan{(A+B)}=\dfrac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}$
Use the identity above with $A=135^o$ and $B=30^o$ to obtain:
\begin{align*}
\require{cancel}
\tan{165^o}&=\tan{(135^o+30^o)}\\\\
&=\frac{\tan{135^o}+\tan{30^o}}{1-\tan{135^o}\tan{30^o}}\\\\
&=\frac{-1+\frac{\sqrt3}{3}}{1-(-1)\left(\frac{\sqrt3}{3}\right)}\\\\
&=\frac{-\frac{3}{3}+\frac{\sqrt3}{3}}{1+\left(\frac{\sqrt3}{3}\right)}\\\\
&=\frac{\frac{-3+\sqrt3}{3}}{\frac{3}{3}+\left(\frac{\sqrt3}{3}\right)}\\\\
&=\frac{\frac{-3+\sqrt3}{3}}{\frac{3+\sqrt3}{3}}\\\\
&=\frac{-3+\sqrt3}{3} \cdot \frac{3}{3+\sqrt3}\\\\
&=\frac{-3+\sqrt3}{\cancel{3}} \cdot \frac{\cancel{3}}{3+\sqrt3}\\\\
&=\frac{-3+\sqrt3}{3+\sqrt3}
\end{align*}
Rationalize the denominator by multiplying $3-\sqrt3$ to both the numerator and denominator to obtain:
\begin{align*}
\require{cancel}
\tan{165^o} &=\frac{-3+\sqrt3}{3+\sqrt3} \cdot \frac{3-\sqrt3}{3-\sqrt3}\\\\
&=\frac{(-3+\sqrt3)(3-\sqrt3)}{(3^2)-\left(\sqrt3\right)^2} &\text{(use the rule }(a-b)(a+b)=a^2-b^2)\\\\
&=\frac{-3(3-\sqrt3)+\sqrt3(3-\sqrt3)}{9-3}\\\\
&=\frac{-9+3\sqrt3+3\sqrt3-3}{6}\\\\
&=\frac{-12+6\sqrt3}{6}\\\\
&=\frac{6(-2+\sqrt3)}{6}\\\
&=\frac{\cancel{6}(-2+\sqrt3)}{\cancel{6}}\\\\
&=-2+\sqrt3
\end{align*}
Thus, $\tan{165^o}=\tan{(135^o+30^o)}=-2+\sqrt3$
