JAMB 1990 · UME · Q39

If $\cos\theta = \frac{12}{13}$ , find $1 + \cot^2\theta$ .

Worked solution (try it first)

Draw a right-angled triangle with adjacent side 12 and hypotenuse 13.
Pythagoras gives the opposite side: $\sqrt{169 - 144} = 5$ .
So $\sin\theta = \frac{5}{13}$ .
Use the identity $1 + \cot^2\theta = \text{cosec}^2\theta$
$= \dfrac{1}{\sin^2\theta}$ , which is $\dfrac{169}{25}$ .
That is option A.