Prove that: (cos theta + cos beta)/(sin beta - sin theta) = (sin theta + sin beta)/(cos theta - cos beta)
{cos(t) + cos(b)} / {sin(b) – sin(t)} = {sin(t) + sin(b)} / {cos(t) – cos(b)}
Now cross-multiply, giving
cos^2(t) – cos^2(b) = -sin*2(t) + sin^2(b)
cos^2(t) + sin^2(t) = cos^2(b) + sin^2(b)
1 = 1
Hence, the original expression is an identity, and LHS = RHS