Hello!
This is a problem about relating values of the Unit Circle.
First, we need to figure out the specific "points" of each angle measure of 30 and 60.
For the angle measure 30, the point will be [tex](\frac{\sqrt{3}}{2},\frac{1}{2})[/tex].
For the angle measure 60, the point will be [tex](\frac{1}{2},\frac{\sqrt{3}}{2})[/tex].
The tangent value of a point will be its [tex]y[/tex] value over its [tex]x[/tex] value.
[tex]\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{2}*\frac{2}{\sqrt{3}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}[/tex]
The cotangent value of a point will be the reciprocal of the tangent value, which we found previously.
[tex]\frac{3}{\sqrt{3}}=\frac{3\sqrt{3}}{3}=\sqrt{3}[/tex]
The sine value of a point will be the [tex]y[/tex] value of the point.
[tex]\frac{\sqrt{3}}{2}[/tex]
So the values we have so far are [tex]\frac{\sqrt{3}}{3}+\sqrt{3}+\frac{\sqrt{3}}{2}[/tex]
Now we have to find the LCD to add these together, which in this case would be 6.
[tex]\frac{2\sqrt{3}}{6}+\frac{6\sqrt{3}}{6}+\frac{3\sqrt{3}}{6}[/tex]
Which adds up to [tex]\frac{11\sqrt{3}}{6}[/tex], which is in simplest radical form.
Hope this helps!
