Respuesta :

LammettHash
[tex]\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}[/tex]

So,

[tex]\tan(135^\circ+30^\circ)=\dfrac{\tan135^\circ+\tan30^\circ}{1-\tan135^\circ\tan30^\circ}[/tex]

You should know that [tex]\tan135^\circ=-1[/tex] and [tex]\tan30^\circ=\dfrac1{\sqrt3}[/tex]

Putting everything together,

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

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