solving $\sec (3 \beta + 10) = \csc (\beta + 8)$

$\sec (3 \beta + 10) = \csc (\beta + 8)$ (in degrees)

I am supposed to find one solution, and the angles are acute.

I do not know the answer or how to get the answer.

It is confusing for me because I don't know what to do with the $3 \beta$.

I am guessing that maybe I get rid of the $\sec$ and $\csc$ somehow but I am not sure how I would go about doing that.

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@Adam: I wonder about how often you are asking these homework questions. So I will hold back answering for a while to allow you proper time to work on it. As far as I can tell, you have asked over 6 today, and over 10 in the last few days. –  mixedmath Jun 6 '11 at 2:01
This is essentially the same problem as your previous question, and can be solved in exactly the same way. The point of posting questions and reading the answers is to try to learn something from the process. –  Arturo Magidin Jun 6 '11 at 2:02
Well I don't know what to do, all I learned from the last one is the change the sign. –  Adam Jun 6 '11 at 2:04
@Adam: Then you didn't learn the right thing. It's exactly the same thing; secant and cosecant are "function and cofunction", to use your parlance. There is one case in which each function and its cofunction are always equal. That gives the answer. –  Arturo Magidin Jun 6 '11 at 2:05
You seem to have a lot of attitude problems and take offense when I don't understand what you say. I don't know why this is, but I am being honest. Most of what you say does not at all further my understand of the material at all. –  Adam Jun 6 '11 at 2:23

Yes. That yields $\csc(26) = \sec(64)$. (degrees) –  ncmathsadist Jun 6 '11 at 2:35