WAEC 2019 · Paper 2 · Q4

(a)
If $\sin X = \dfrac{p - q}{p + q}$ , where $0^\circ \le X \le 90^\circ$ , find $1 - \tan^2 X$ .

Worked solution (try it first)

Draw a right-angled triangle with opposite side $p - q$ and hypotenuse $p + q$ .
Adjacent side: $\sqrt{(p + q)^2 - (p - q)^2} = \sqrt{4pq}$
$= 2\sqrt{pq}$ .
So $\tan X = \dfrac{p - q}{2\sqrt{pq}}$ and $\tan^2 X = \dfrac{p^2 - 2pq + q^2}{4pq}$ .
$1 - \tan^2 X = \dfrac{4pq - p^2 + 2pq - q^2}{4pq}$
$= \dfrac{6pq - p^2 - q^2}{4pq}$ .