# Finding $\csc \theta$ given $\cot \theta$

I have the following problem:

If $\cot{C} = \frac{\sqrt{3}}{7}$, find $\csc{C}$

From my trig identities, I know that $\cot{\theta} = \frac{1}{\tan{\theta}}$, and $\csc{\theta} = \frac{1}{\sin{\theta}}$, and also $\cot{\theta} = \frac{\cos{\theta}}{\sin{\theta}}$

However, I can't seem to see how to connect the dots to get from cotangent to cosecant. I figure I might be able to use the last identity if I can somehow make $\cos{C} = 1$, but I don't really see how to do that, either.

This is homework, so please provide me with some pointers rather than complete solutions.

Thanks.

In my answer here, I describe some general ideas for how to use one trig function of an angle to determine another trig function of the same angle. The basic idea is to draw triangle(s) on a coordinate system that correspond to the given trig function and use those to compute the other trig function.

• Thanks, I'll try drawing some triangles and see if it helps. Mar 7, 2011 at 6:31
• @friedo: the diagrams in that other answer might also be helpful. Mar 7, 2011 at 6:33

From $\sin^2\theta + \cos^2\theta = 1$, divide through by $\sin^2\theta$ to get a relation between $\cot^2\theta$ and $\csc^2\theta$.

P.S. The information given is not enough, though, to determine the value of $\csc\theta$ unless you happen to know which quadrant you are working in; you know that you are in either quadrant I or III, since the cotangent is positive; but that does not tell you whether the cosecant is positive or negative; you'll have two possible answers. This is pretty much the same situation as how, if you know that $\sin\theta=\frac{1}{2}$, this only determines $\cos\theta$ up to sign.

• Thanks, I think this will help. I do know that we're working in quadrant I since we're still on right triangles. Mar 7, 2011 at 6:30
• @friedo: In that case, Isaac's solution is very useful. Just draw any triangle with the right cotangent (the easiest is one in which the adjacent side is the numerator you have, and the opposite side is the denominator); use Pythagoras's Theorem to compute the hypothenuse, and then just read off the value of the cosecant. It is, really, the same as this method (you are using the corresponding identity when you compute the value of the hypothenuse) but it's easier to visualize. Mar 7, 2011 at 6:33

Here's a simple way I usually think about it. Suppose you have a right triangle in the first quadrant. Since $\cot C=\frac{\sqrt{3}}{7}$, you know the ratio of the leg adjacent to $C$ to the leg opposite of $C$ is $\frac{\sqrt{3}}{7}$. So let's just say the opposite leg has length $7$ and the adjacent leg has length $\sqrt{3}$. Then by the Pythagorean theorem, the hypotenuse has length $\sqrt{52}=2\sqrt{13}$.

Now $\csc C$ is just the ratio of the hypotenuse to the opposite leg, essentially the inverse of $\sin$, which is the ratio of the opposite leg to the hypotenuse.