JAMB 1987 · UME · Q39

If $\tan\theta = \dfrac{m^2 - n^2}{2mn}$ , find $\sec\theta$ .

Worked solution (try it first)

Draw a right-angled triangle with the opposite side $m^2 - n^2$ and the adjacent side $2mn$ .
Pythagoras: the hypotenuse squared is $(m^2 - n^2)^2 + 4m^2n^2 = m^4 + 2m^2n^2 + n^4$
$= (m^2 + n^2)^2$ , so the hypotenuse is $m^2 + n^2$ .
$\sec\theta = \dfrac{\text{hypotenuse}}{\text{adjacent}}$
$= \dfrac{m^2 + n^2}{2mn}$ , option B.