JAMB 2002 · UME · Q13

If $\tan\theta = \frac43$ , calculate $\sin^2\theta - \cos^2\theta$ .

Worked solution (try it first)

$\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$
$= \frac43$ .
Pythagoras gives the hypotenuse $\sqrt{16 + 9} = 5$ .
So $\sin\theta = \frac45$ and $\cos\theta = \frac35$ .
Then $\sin^2\theta - \cos^2\theta = \frac{16}{25} - \frac{9}{25}$
$= \frac{7}{25}$ , option A.