# Question c4250

Jun 26, 2016

By definition, $\sec \theta = \frac{1}{\cos} \theta$.

#### Explanation:

Hence, $\cos \theta = - \frac{4}{5}$. We know that $\tan \theta > 0$, and the only quadrant where tangent is positive and the other trig ratios are negative is quadrant III.

Since we only know the side adjacent $\theta$ and the hypotenuse, we must find the side opposite $\theta$.

We can do this using Pythagorean theorem. Let a be $- 4$ and $c$ be 5.

${a}^{2} + {b}^{2} = {c}^{2}$

${b}^{2} = {c}^{2} - {a}^{2}$

${b}^{2} = {\left(5\right)}^{2} - {\left(- 4\right)}^{2}$

${b}^{2} = 25 - 16$

$b = \sqrt{9}$

$b = \pm 3$

We will take the $- 3$, because in quadrant three both opposite and adjacent sides to the angle $\theta$ will be negative.

Now that we know that

$\text{adjacent = -4}$
$\text{opposite} = - 3$
$\text{hypotenuse = 5}$

We can define cosine and cotangent. Cosine is adjacent/hypotenuse, and cotangent is 1/tantheta = 1/("opposite"/"adjacent") = "adjacent"/"opposite"#.

Applying these definitions to the problem at hand, we have:

$\cot \theta = \frac{4}{3}$

$\cos \theta = - \frac{4}{5}$

Now, adding these is simple arithmetic.

$\frac{4}{3} + \left(- \frac{4}{5}\right) = \frac{20}{15} - \frac{12}{15} = \frac{8}{15}$

Thus, $\cos \theta + \cot \theta = \frac{8}{15}$.

Hopefully this helps!