# Finding $\csc$ with $\cot$

I know that $\cot\theta = 4/3$ how do I find $\csc\theta$?

I tried to do $\csc^2\theta - \cot^2\theta = 1$

This gives me $\csc^2\theta = 1 + \cot^2\theta$

this gives $csc^2\theta = 9/9 + 16/9 = 27/9 = \sqrt{3}$

is this wrong? My book gives the answer as $5/3$

I can never go more than $2$ homework problems without getting stuck.

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You did everthing right, but 9+16 = 25 not 27 –  crasic Jun 28 '11 at 22:02

I don't see what is your problem here... You know that $csc^2 x= 1+\cot^2 x=1+\frac{16}{9}=\frac{25}{9}$. From here, you get $\csc x=\frac{5}{3}$. It's pure algebra. Just look at what you have and where do you want to get, and as I said, do not rush with computations, since I notice you make very many elementary mistakes.

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Yeah I checked my work about 12 times, I am attempting to review 14 sections before tomorrow to catch up since I am so far behind in class. I will likely get two or three done at this rate. –  Adam Jun 28 '11 at 22:06
Well, since you had time to check 12 times, try and slow down your computation and writing speed, and think very well every step. Don't skip anything as being too easy. –  Beni Bogosel Jun 28 '11 at 22:10
@Adam as I pointed out in my comment, even though you checked 12 times you missed the clear arithmetic error –  crasic Jun 28 '11 at 22:13

If $\cot{(x)} = 4/3$, then we have this picture

<

                 /|
/ |  3
5   /  |
/   |
x-----
4


Now compute.

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It would help a bit more simply by being clear about which angle is $\theta$, so as not to confuse Adam...I just added the angle "x" to illustrate your relation. –  amWhy Jun 28 '11 at 22:50
@Adam trig functions represent relations between the sides of a right triangle. Excepting some corner cases (like $\theta = \frac{\pi}{2}$) you can always construct a right triangle that "represents" the trig relation. –  crasic Jun 28 '11 at 22:15