Simple Trig Trio Problems - I keep getting the wrong answer

[jxcov]

Okay, I have solved this many times and I am getting a different answer from the teacher key.
Given cosA = 4/5 and tanB = 5/12 with both angles in quadrant one, find the exact value of the given function.
14) sin(A-B)
15) sec(A+B)
16) cot(B-A)

So! Here is how I started solving....

given the cosA = 4/5 I found the triangle values, 5 hypotenuse, 4 Adjacent, 4 Opposite
given the tanB = 5/12; the hypotenuse is 13, 12 adjacent, 5 opposite...Right? They should be Pythagorean triples.

By taking that, for #14 I plugged in
sin ((3/5) - (5/13))
which led me to
sin((39/65) - (25/13)) = sin(14/65)
Is that not the answer? The key says: 16/65. Did I make a mistake?

Now for #15 sec(A+B)
take them same numbers and shouldn't it be sec((5/4) + (13/12))?
by reducing that I get: 7/3 but the answer key says 65/33 what did I do wrong? Isn't cos = Adjacent / Hypotenuse making sec = Hypotenuse / Adjacent

for #16 cot(B-A)
((12/5) - (4/3)) simplified to
(16/15) while the key says -63/16

Thank you so much for help, sorry if my formatting is bad. //edit: formatting

 

[ksdhart2]

You're kind of on the right track, but it looks like everything's getting a bit muddled in your head. There's two main ways you can solve this problem. One is, I believe, the more straightforward and easier solution, but requires a calculator; the other is a bit more obtuse, but the calculations can be done without a calculator when you're given "nice" values like you were here (nice for a test where you might not have access to a calculator). I'll start with the more straightforward one.

You're given the values of cos(A) = 4/5 and tan(B) = 9/12. You can use these to directly solve for A and B. Namely, A = cos^-1(4/5) and B=tan^-1(9/12). Due to the nature of inverse trig functions, these will be the angle measures of A and B. You can then plug these values into the equations, like so: sin(A-B)=sin(cos^-1(4/5)+tan^-1(9/12)).

The slightly longer way is very much like what you've done. You're given cos(A) = 4/5. Using the mnemonic SOHCAHTOA, you then found the associated values of the right triangle. Namely "5 hypotenuse, 4 Adjacent, 3 Opposite" (I corrected what I assume is a typo here, considering you used the correct value of 3 later on). Up to here, your working is correct, but it's after this point you've gone off the rails. It appears to me as if you've gotten sin(A)=opposite/hypotenuse=3/5 confused with the (as of yet unknown) angle measure of A. These two values are more definitely not the same. Be sure you understand why and the difference between them.

You do know the value of cos(A) and sin(A) though, as well as the values of sin(B) and cos(B) you can derive from correcting the process with the other triangle for B. So you can use one of the known identities, specifically the angle addition and subtractionhttp://www.cut-the-knot.org/triangle/SinCosFormula.shtml formulas you learned:

sin(A - B) = [sin(A) * cos(B)] - [cos(A) * sin(B)]

The other two problems are also solvable in the same manner, if you note that sec(A + B) = 1/cos(A + B), and cot(B - A)=cos(B - A)/sin(B - A)

 

[jxcov]

(I corrected what I assume is a typo here, considering you used the correct value of 3 later on).

"Oops"
I can't thank you enough @ksdhart2, I had just solved 10 similar problems using the sum & difference identities but the sec and cot ones threw me off as I am positive the teacher went over them, (while it is just 1/cos identity LOL, I'll take fault).

I see what you mean about the sin(A) with the angle measurement confusion I had there. So for the sin(A-B) I ended up getting ( (3/5) * (5/13)) + ( (3/5) * (5/13)) and that gives me 16/65, the right answer!
I have done the rest on paper, I promise but I am too lazy to type out.

Again, thank you so much! It clicked in my head!