Question

Sum and difference tan(135+30)

Answer 1

$\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\sin y\cos x}{\cos x\cos y-\sin x\sin y}=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$

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,

$\tan(135^\circ+30^\circ)=\dfrac{-1+\dfrac1{\sqrt3}}{1-(-1)\left(\dfrac1{\sqrt3}\right)}=\dfrac{1-\sqrt3}{\sqrt3+1}=\sqrt3-2$