Answer
Both sides of the equation for the Law of Tangents are equal, which verifies this law for the information given in the question.
Work Step by Step
$a = 2$
$b = 2\sqrt{3}$
$A = 30^{\circ}$
$B = 60^{\circ}$
We can find the value of the left side of the equation for the Law of Tangents:
$\frac{tan~\frac{1}{2}(A-B)}{tan~\frac{1}{2}(A+B)} = \frac{tan~\frac{1}{2}(30^{\circ}-60^{\circ})}{tan~\frac{1}{2}(30^{\circ}+60^{\circ})}$
$\frac{tan~\frac{1}{2}(A-B)}{tan~\frac{1}{2}(A+B)} = \frac{tan~\frac{1}{2}(-30^{\circ})}{tan~\frac{1}{2}(90^{\circ})}$
$\frac{tan~\frac{1}{2}(A-B)}{tan~\frac{1}{2}(A+B)} = \frac{tan~(-15^{\circ})}{tan~(45^{\circ})}$
$\frac{tan~\frac{1}{2}(A-B)}{tan~\frac{1}{2}(A+B)} = -0.2679$
We can find the value of the right side of the equation for the Law of Tangents:
$\frac{a-b}{a+b} = \frac{2-2\sqrt{3}}{2+2\sqrt{3}} = -0.2679$
Both sides of the equation for the Law of Tangents are equal, which verifies this law for the information given in the question.
