It was reported that majority of the candidates found this question quite challenging. Only a few candidates were able to express tan 30° in the form 2tan15.
1 – tan215
Quite a number of them expressed tan15° in the form tan 45° - tan 30°
1 + tan 30° tan 45°
They were expected to substitute 15° for A in the expression to get
tan 2(15) = tan 30 = 1/√3 = 2tan 15° and simplify to obtain
1-tan215°
tan215° + 2√3tan 15° - 1 = 0, a quadratic equation in tan 15°. Solving this equation gives tan 15° = +(2 - ) since tan 15° is always positive, the correct answer was tan 15° = 2 - √3.
