Given that $\arctan \left( {\frac{1}{5}} \right) + \arctan \left( {\frac{1}{8}} \right) = \arctan \left( {\frac{1}{p}} \right)$, where $p \in {\mathbb{Z}^ + }$, find p.

Hence find the value of $\arctan \left( {\frac{1}{2}} \right) + \arctan \left( {\frac{1}{5}} \right) + \arctan \left( {\frac{1}{8}} \right)$.

attempt at use of $\tan (A + B) = \frac{{\tan (A) + \tan (B)}}{{1 - \tan (A)\tan (B)}}$     M1

$\frac{1}{p} = \frac{{\frac{1}{5} + \frac{1}{8}}}{{1 - \frac{1}{5} \times \frac{1}{8}}}{\text{ }}\left( { = \frac{1}{3}} \right)$     A1

$p = 3$     A1

Note: the value of p needs to be stated for the final mark.

[3 marks]

$\tan \left( {\arctan \left( {\frac{1}{2}} \right) + \arctan \left( {\frac{1}{5}} \right) + \arctan \left( {\frac{1}{8}} \right)} \right) = \frac{{\frac{1}{2} + \frac{1}{3}}}{{1 - \frac{1}{2} \times \frac{1}{3}}} = 1$     M1A1

$\arctan \left( {\frac{1}{2}} \right) + \arctan \left( {\frac{1}{5}} \right) + \arctan \left( {\frac{1}{8}} \right) = \frac{\pi }{4}$     A1

[3 marks]

Those candidates who used the addition formula for the tangent were usually successful.

Some candidates left their answer as the tangent of an angle, rather than the angle itself.