If sintheta=3/5, what is costheta + cottheta?

$\cos \theta + \cot \theta = \text{adj"/"hyp"+"adj"/"opp} = \frac{4}{5} + \frac{4}{3} = \frac{12}{15} + \frac{20}{15} = \frac{32}{15}$ in Q1. $- \frac{32}{15}$ in Q2.

Explanation:

If $\sin \theta = \frac{3}{5}$, what is $\cos \theta + \cot \theta$?

First, let's remember that $\sin \theta$ is the ratio of the opposite side divided by the hypotenuse:

$\sin = \text{opp"/"hyp}$

So we know with this triangle that we have a side = 3 and hypotenuse = 5.

There is a special kind of right triangle called the "3, 4, 5 triangle", so called because of the ease in calculating sides using the pythagorean theorem:

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

And so if we didn't know or remember the "3, 4, 5 triangle", we could find it this way:

${3}^{2} + {b}^{2} = {5}^{2}$

$9 + {b}^{2} = 25$

${b}^{2} = 16$

$b = 4$

With opposite = 3, adjacent = 4, and hypotenuse = 5, we can answer the question.

Ok, so $\sin \theta$ is positive, which means the opposite is positive (the hypotenuse is always positive). The quadrants on the cartesian chart where the opposite is positive (i.e. has a positive y-value and not a negative one) is in Q1 and Q2. So let's solve for each quadrant:

For Q1:
$\cos \theta + \cot \theta = \text{adj"/"hyp"+"adj"/"opp} = \frac{4}{5} + \frac{4}{3} = \frac{12}{15} + \frac{20}{15} = \frac{32}{15}$

For Q2:
$\cos \theta + \cot \theta = \text{adj"/"hyp"+"adj"/"opp} = - \frac{4}{5} - \frac{4}{3} = - \frac{12}{15} - \frac{20}{15} = - \frac{32}{15}$