# solving $\sec (3 \beta + 10) = \csc (\beta + 8)$ [duplicate]

$\sec (3 \beta + 10) = \csc (\beta + 8)$ (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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## marked as duplicate by Joel Reyes Noche, Mark Fantini, N. F. Taussig, Jack D'Aurizio, LutzLMar 26 at 14:28

@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
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
The same question was added to this question of yours math.stackexchange.com/questions/43456/angle-equations . Why? –  Américo Tavares Jun 6 '11 at 10:44

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