(a)     Show that $\arctan \left( {\frac{1}{2}} \right) + \arctan \left( {\frac{1}{3}} \right) = \frac{\pi }{4}$ .

(b)     Hence, or otherwise, find the value of $\arctan (2) + \arctan (3)$ .

(a)     METHOD 1

let $x = \arctan \frac{1}{2} \Rightarrow \tan x = \frac{1}{2}$ and $y = \arctan \frac{1}{3} \Rightarrow \tan y = \frac{1}{3}$

$\tan (x + y) = \frac{{\tan x + \tan y}}{{1 - \tan x\tan y}} = \frac{{\frac{1}{2} + \frac{1}{3}}}{{1 - \frac{1}{2} \times \frac{1}{3}}} = 1$     M1

so, $x + y = \arctan 1 = \frac{\pi }{4}$     A1AG

METHOD 2

for $x,{\text{ }}y > 0$ , $\arctan x + \arctan y = \arctan \left( {\frac{{x + y}}{{1 - xy}}} \right)$ if $xy < 1$     M1

so, $\arctan \frac{1}{2} + \arctan \frac{1}{3} = \arctan \left( {\frac{{\frac{1}{2} + \frac{1}{3}}}{{1 - \frac{1}{2} \times \frac{1}{3}}}} \right) = \frac{\pi }{4}$     A1AG

METHOD 3

an appropriate sketch     M1

e.g.    

correct reasoning leading to $\frac{\pi }{4}$     R1AG

(b)     METHOD 1

$\arctan (2) + \arctan (3) = \frac{\pi }{2} - \arctan \left( {\frac{1}{2}} \right) + \frac{\pi }{2} - \arctan \left( {\frac{1}{3}} \right)$     (M1)

$= \pi - \left( {\arctan \left( {\frac{1}{2}} \right) + \arctan \left( {\frac{1}{3}} \right)} \right)$     (A1)

Note: Only one of the previous two marks may be implied.

$= \pi - \frac{\pi }{4} = \frac{{3\pi }}{4}$     A1     N1

METHOD 2

let $x = \arctan 2 \Rightarrow \tan x = 2$ and $y = \arctan 3 \Rightarrow \tan y = 3$

$\tan (x + y) = \frac{{\tan x + \tan y}}{{1 - \tan x\tan y}} = \frac{{2 + 3}}{{1 - 2 \times 3}} = - 1$     (M1)

as $\frac{\pi }{4} < x < \frac{\pi }{2}\,\,\,\,\,\left( {{\text{accept }}0 < x < \frac{\pi }{2}} \right)$

and $\frac{\pi }{4} < y < \frac{\pi }{2}\,\,\,\,\,\left( {{\text{accept }}0 < y < \frac{\pi }{2}} \right)$

$\frac{\pi }{2} < x + y < \pi \,\,\,\,\,{\text{(accept }}0 < x + y < \pi )$     (R1)

Note: Only one of the previous two marks may be implied.

so, $x + y = \frac{{3\pi }}{4}$     A1     N1

METHOD 3

for $x,{\text{ }}y > 0$ , $\arctan x + \arctan y = \arctan \left( {\frac{{x + y}}{{1 - xy}}} \right) + \pi {\text{ if }}xy > 1$     (M1)

so, $\arctan 2 + \arctan 3 = \arctan \left( {\frac{{2 + 3}}{{1 - 2 \times 3}}} \right) + \pi$     (A1)

Note: Only one of the previous two marks may be implied.

$= \frac{{3\pi }}{4}$     A1     N1

METHOD 4

an appropriate sketch     M1

e.g.    

correct reasoning leading to $\frac{{3\pi }}{4}$     R1A1

[5 marks]

Most candidates had difficulties with this question due to a number of misconceptions, including $\arctan x = {\tan ^{ - 1}}x = \frac{{\cos x}}{{\sin x}}$ and $\arctan x = \frac{{\arcsin x}}{{\arccos x}}$, showing that, although candidates were familiar with the notation, they did not understand its meaning. Part (a) was done well among candidates who recognized arctan as the inverse of the tangent function but just a few were able to identify the relationship between parts (a) and (b). Very few candidates attempted a geometrical approach to this question.