Finding cot(a+2B) Given tan(a) and Sin(b)

[GrandMa5TR]

The full question reads: Suppose tan a = -3/4 and sin B = 12/12 where a is a second quadrant is angle and B is a first quadrant angle. Find cot(a+2B)

Any help is appreciated, but I'd be especially grateful if you could show me how you worked it out.

 

[Deleted member 4993]

The full question reads: Suppose tan a = -3/4 and sin B = 12/12 where a is a second quadrant is angle and B is a first quadrant angle. Find cot(a+2B)

Any help is appreciated, but I'd be especially grateful if you could show me how you worked it out.

What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33

 

[GrandMa5TR]

What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/announcement.php?f=33

Well I know that tan a = -3/4 is just arctan, and the same thing for sine. So I tried to make two triangles using both tan and sin. I then used some identies to find cota = -3/4 and cot2B = -120/119. I then considered using cot(a+B) = ((cot a)(cot B)-1)/(cotB-cota). But I don't think it would work here. So that was my crack at it, but now I'm stumped. What are your thoughts on the problem?