Let $\theta$ be an angle in quadrant IV such that $\sin \theta = −12/13$.
Then, find the exact values of $\sec\theta$ and $\cot\theta$.
I've done the Pythagorean theorem: 5 for the adjacent side.
$\sec\theta = -12/5$
$\cot\theta = -5/13$
Is this correct?
You're right that 5 would be the adjacent side. So we have a right triangle, where the adjacent side is 5, the opposite side is -12, and the hypotenuse is 13.
$\sec \theta = \frac{1}{\cos \theta}$, so $\sec \theta$ is (hypotenuse)/(adjacent) $=\frac{13}{5}$.
$\cot \theta = \frac{1}{\tan \theta}$, so $\sec \theta$ is (adjacent)/(opposite) $=-\frac{5}{12}$.
Alternatively, note that $\sin ^2 \theta + \cos^2 \theta = 1$, so $(-\frac{12}{13})^2 + \cos^2 \theta = 1$, and $\cos \theta = \pm \frac{5}{13}$. Since we're in quadrant IV, we know that $\cos \theta$ is positive, so $\cos \theta = \frac{5}{13}$.
Then $\sec \theta = \frac{1}{\cos \theta} = \frac{13}{5}$.
And $\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta} = -\frac{5}{12}$.
You have worked out that this is a $5 - 12 - 13$ triangle. With 5 as the base, 12 as the height and 13 as the hypotenuse.
$\sec \theta = \frac 1{\cos\theta} = \frac {\text{Adjacent}}{\text {Hypotenuse}}$
$\cot \theta = \frac {\cos\theta}{\sin\theta}$
Next question. What are the signs? In Q IV. $\cos \theta > 0, \sin \theta < 0$
What do you think the signs of $\sec\theta, \cot \theta?$
I think you have made mistake with sides.
$\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} $
So comparing with values,
Perpendicular = 12 and Hypotenuse = 13
Using phythagorous you got base as 5.
$\cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} $
$\cos \theta = \frac{5}{13}$
As $\cos\theta$ positive as in IV quadrant.
$\cot \theta = \frac{\text{Base}}{\text{Perpendicular}} $
$\cot \theta = \frac{-5}{12}$
As $\cot\theta$ negative as in IV quadrant.