Suppose sin A = 12/13 with 90º≤A≤180º. Suppose also that sin B = -7/25 with -90º≤B≤0º. Find tan (A – B).

Answer 1

sin^-1(12/13) = 67.38

180-67.39 = 112.62 = A

sin^-1(-7/25) = -16.26 = B

Tan ( A - B) = ???

I think that's how you approach it.. someone verify please?

Answer 2

I'd do it like this:

in QII if sinA = 12/13, cosA = -5/13
in QIV, if sinB = -7/25, cosB = 24/25

So,

tanA = -12/5
tanB = -7/24

tan(A-B) = ((-12/5)-(-7/24))/(1+(-12/5)(-7/24))
= (-253/120) / (204/120)
= -253/204

Answer 3

Thank you guys! :)That really helped! I understand now!