#### Answer

The product of the diagonals is equal to the sum of the products of the opposite sides. Therefore:
$sin(A+B) = sin~A~cos~B+sin~B~cos~A$
Note that this is the formula for the sine of the sum of two angles.

#### Work Step by Step

The line through the middle of the circle is the diameter. We know that any triangle constructed with the diameter as one side and with a point on the circle's circumference is a right triangle. Therefore, the four sides around the outside have lengths $sin~A, cos~A, sin~B,$ and $cos~B$ since the hypotenuse has a length of 1.
In Exercise 53, we saw that for any triangle inscribed in a circle, $\frac{d}{sin~D} = 2r$. In this case, let $d$ be the side opposite the angle $A+B$. Then:
$\frac{d}{sin~(A+B)} = 2r = 1$
$d = sin~(A+B)$
Therefore, the side opposite the angle $A+B$ has a length of $sin(A+B)$
The product of the diagonals is $sin~(A+B)(1) = sin(A+B)$
The product of one pair of opposite sides is $sin~A~cos~B$
The product of the other pair of opposite sides is $sin~B~cos~A$
The sum of these two products is $sin~A~cos~B+sin~B~cos~A$
From the information in the question, the product of the diagonals is equal to the sum of the products of the opposite sides. Therefore:
$sin(A+B) = sin~A~cos~B+sin~B~cos~A$
Note that this is the formula for the sine of the sum of two angles.