# 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

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– 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

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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
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## 2 Answers

These two functions are cofunctions. Using this fact gives you one easy solution.

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I don't quite see what to do, is this correct? So 4b+18 = 90 so b=18? – Adam Jun 6 '11 at 2:32
Yes.  That yields $\csc(26) = \sec(64)$.  (degrees) – ncmathsadist Jun 6 '11 at 2:35
Awesome so if it adds to 90 I know it is correct. – Adam Jun 6 '11 at 2:37

look at this site

http://en.wikipedia.org/wiki/Trigonometric_functions here is explanation what sec and csc are

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