Recall the angle sum identities:
[tex]\sin(x+y)=\sin x\cos y+\cos x\sin y[/tex]
[tex]\cos(x+y)=\cos x\cos y-\sin x\sin y[/tex]
Now,
[tex]\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}[/tex]
Divide through numerator and denominator by [tex]\cos x\cos y[/tex] to get
[tex]\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}[/tex]
Next, we use the fact that [tex]x,y[/tex] lie in the first quadrant to determine that
[tex]\sin x=\dfrac12\implies\cos x=\sqrt{1-\sin^2x}=\dfrac{\sqrt3}2[/tex]
[tex]\cos y=\dfrac{\sqrt2}2\implies\sin x=\sqrt{1-\cos^2x}=\dfrac1{\sqrt2}[/tex]
So we then have
[tex]\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}[/tex]
[tex]\tan y=\dfrac{\sin y}{\cos y}=\dfrac{\frac1{\sqrt2}}{\frac{\sqrt2}2}=1[/tex]
Finally,
[tex]\tan(x+y)=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3\approx3.73[/tex]
