# Suppose in a right triangle, cos(t)=3/4. How do you find: cot(t)?

Jun 20, 2018

$\cot \left(t\right) = \frac{3}{\sqrt{7}}$

#### Explanation:

Starting from the fundamental equation

${\sin}^{2} \left(t\right) + {\cos}^{2} \left(t\right) = 1$

we can deduce

sin(t) = \pm\sqrt(1-cos^2(t)

So, in our case,

$\sin \left(t\right) = \setminus \pm \setminus \sqrt{1 - \frac{9}{16}} = \setminus \pm \setminus \sqrt{\frac{7}{16}}$

Actually, we have $\sin \left(t\right) = \setminus \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$

In fact, since $t$ is an an angle of a right triangle, it must be acute: the inner angles of a triangle sum to $\pi$, but since one angle is right, the other two must sum to $\frac{\pi}{2}$, and thus they are acute.

An acute angle lies in the first quadrant, where both sine and cosine are positive, hence the choice of the sign.

Now that we know both sine and cosine, we can compute the cotangent by its definition:

$\cot \left(t\right) = \cos \frac{t}{\sin} \left(t\right) = \frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}} = \frac{3}{\cancel{4}} \cdot \frac{\cancel{4}}{\sqrt{7}} = \frac{3}{\sqrt{7}}$