Question

If cos p = - 4/5 with p in quadrant 3,
and cos q = 5/13 with q in quadrant 4,
find tan (p - q)

A. 1
B. -63/16
c. 63/16
d. - 33/16

Answer 1

If $\cos p = -4/5$ with $p$ in quadrant 3, and $\cos q = 5/13$ with $q$ in quadrant 4, find $\tan(p - q)$

a. 1

b. -63/16

c. 63/16

d. -33/16

**Solution.**

$$\tan(p - q) = \frac{\tan p - \tan q}{1 + \tan p \cdot \tan q}$$ $$\tan p = \frac{\sin p}{\cos p}, \tan q = \frac{\sin q}{\cos q}$$ $$\sin^2 p + \cos^2 p = 1,$$

because $p$ is in quadrant 3 $\sin p = -\sqrt{1 - \cos^2 p}$, $\sin p = -\sqrt{1 - \frac{16}{25}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$.

$$\sin^2 q + \cos^2 q = 1,$$

because $q$ is in quadrant 4 $\sin q = -\sqrt{1 - \cos^2 q}$, $\sin p = -\sqrt{1 - \frac{25}{169}} = -\sqrt{\frac{144}{169}} = -\frac{12}{13}$.

$$\tan p = -\frac{3}{5} \div \left(-\frac{4}{5}\right) = \frac{3}{4}.$$ $$\tan q = -\frac{12}{13} \div \frac{5}{13} = -\frac{12}{5}.$$ $$\tan(p - q) = \frac{\frac{3}{4} + \frac{12}{5}}{1 - \frac{3}{4} \cdot \frac{12}{5}} = -\frac{63}{16}.$$

**Answer.**

b. -63/16