WAEC 2012 · Paper 2 · Q1

(a)
Simplify: $\dfrac{1\frac14 + \frac79}{1\frac49 - 2\frac23 \times \frac{9}{64}}$ .
(b)
Given that $\sin x = \frac23$ , evaluate, leaving your answer in surd form and without using tables or a calculator, $\tan x - \cos x$ .

Worked solution (try it first)

(a)

Work out the top and bottom separately.
Top: $1\frac14 + \frac79 = \frac54 + \frac79$
$= \frac{45 + 28}{36}$
$= \frac{73}{36}$ .
Bottom: multiply first, $2\frac23 \times \frac{9}{64} = \frac83 \times \frac{9}{64}$
$= \frac38$ .
Then $1\frac49 - \frac38 = \frac{13}{9} - \frac38$
$= \frac{104 - 27}{72}$
$= \frac{77}{72}$ .
Divide: $\frac{73}{36} \times \frac{72}{77} = \frac{146}{77}$
$= 1\frac{69}{77}$ .

(b)

$\sin x = \frac23$ : draw a right-angled triangle with opposite 2 and hypotenuse 3.
The adjacent side is $\sqrt{9 - 4} = \sqrt5$ .
So $\cos x = \frac{\sqrt5}{3}$ and $\tan x = \frac{2}{\sqrt5}$ .
Rationalise: $\tan x = \frac{2}{\sqrt5} \times \frac{\sqrt5}{\sqrt5}$
$= \frac{2\sqrt5}{5}$ .
Then $\tan x - \cos x = \frac{2\sqrt5}{5} - \frac{\sqrt5}{3}$
$= \frac{6\sqrt5 - 5\sqrt5}{15}$
$= \frac{\sqrt5}{15}$ .

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