# Basic trig test question

1. Mar 9, 2012

### e^(i Pi)+1=0

1. The problem statement, all variables and given/known data
Given that cot $\theta$ = -12/5 and csc $\theta$ < 0, find sec$\theta$.

This was a question on a test that I drew a blank on, and I'm still not sure how to handle it due to my "teacher" repeatedly dismissing me when I try asking about it. Now, it occurred to me that this could be solved using trig identities and substitution. Starting with $sin^{2}$$\theta$ + $cos^{2}$$\theta$ = 1 and sin/cos = -5/12, I ended up with -
sin = -5/13
cos= 12/13
tan= -5/12
sec= 13/12

and I am confident this is the right answer. But, we have not covered trig identities in class so I am sure there is another easier way to solve this. My question is...what is it? Also, say you take the square root of a trig identity in an equation - how do you know weather it is positive or negative? Thanks in advance.

Last edited: Mar 9, 2012
2. Mar 9, 2012

### tiny-tim

hi e^(i Pi)+1=0!
that is the way

(though it would be easier to memorise and use sec2 = tan2 + 1, csc2 = cot2 + 1 )

(another way of course is to say that if cot = 12/5, then it's obviously a 5,12,13 triangle, and then eg sec will be hyp/adj)
you need to be told (as in this question)

btw, i can't see any latex … are other people having this problem?​

3. Mar 9, 2012

### sjb-2812

Even if you haven't covered $sin^{2}$$\theta$ + $cos^{2}$$\theta$ = 1 formally, I guess you could envisage a right-angled triangle with adjacent 12 and opposite 5, and get the hypotenuse with Pythagoras, for one.

4. Mar 9, 2012

### e^(i Pi)+1=0

Sometimes I don't see latex, but it always pops up after I refresh. Thank you for the quick responses.

edit - actually, my answer WAS wrong since I started with tan = -12/5 when it was -5/12, but it's fixed now.

Last edited: Mar 9, 2012