How do you solve for secant with given cotangent and cosecant?

[e^(i Pi)+1=0]

Homework Statement

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.

 

[tiny-tim]

hi e^(i Pi)+1=0!

… 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?

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)

Also, say you take the square root of a trig identity in an equation - how do you know weather it is positive or negative?

you need to be told (as in this question)

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

 

[sjb-2812]

Homework Statement

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 = -12/5, I ended up with -
sin = -12/13
cos= 5/13
tan= -5/12
csc= -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.

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.

 

[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.