First we need to find `arcsin(-1/2)` and `arcsin(-1).` By memory or a 30-60-90 triangle or a trig table, we know that `sin(-30^o)=-1/2` and `sin(-90^o)=-1,` and `-30` and `-90` are in the range of `arcsin,` which is `[-90,90]` ,so `arcsin(-1/2)` `=-30^o` and `arcsin(-1)=-90^o`

Therefore,

`tan(arcsin(-1/2)+arcsin(-1))=tan(-30^o +(-90^o))=tan(-120^o).` This is where a unit circle will come in handy (the math is being glitchy now if I use the degree symbol, so I'll drop it from now on).

`tan(-120)=sin(-120)/cos(-120)=(-sqrt(3)/2)/(-1/2)=sqrt(3).`

Another way is to use symmetry and get

`tan(-120)=tan(60)=sqrt(3)`. Either way,

`tan(arcsin(-1/2)+arcsin(-1))=sqrt(3).`

**Further Reading**