if (a+b)tan(theta-phi) = (a-b)tan(theta + phi) and a cos2 phi + b cos2 theta =c then prove that a square - b square + c square = 2 ac cos2 phi.
a+b/a-b = tan(t + p)/tan(t-p)
You also have the equation , from which
Square the equations (1) and (2) and add:
That simplifies to give the result that you want.
Notice that this is really a problem about a triangle, with sides a, b, c and angles (opposite the side a) and (opposite the side b). The equation (1) is then the sine rule, and the equation is the cosine rule.