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Further Mathematics Paper 2, May/June. 2009  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
General Comments
Weakness/Remedies
Strength



Question 4

                              2tanA
Given that tan 2A = 1 – tan2A, evaluate tan 15°, leaving your answer in surd form.

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Observation

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.

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