\(\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\sin y\cos x}{\cos x\cos y-\sin x\sin y}=\dfrac{\tan x+\tan y}{1-\tan x\tan y}\)
So,
\(\tan(135^\circ+30^\circ)=\dfrac{\tan135^\circ+\tan30^\circ}{1-\tan135^\circ\tan30^\circ}\)
You should know that \(\tan135^\circ=-1\) and \(\tan30^\circ=\dfrac1{\sqrt3}\)
Putting everything together,
\(\tan(135^\circ+30^\circ)=\dfrac{-1+\dfrac1{\sqrt3}}{1-(-1)\left(\dfrac1{\sqrt3}\right)}=\dfrac{1-\sqrt3}{\sqrt3+1}=\sqrt3-2\)
