We are given: .
We have: .
. . . .
finding the exact value of tan (α-β) when:
tan α = -(5/12) in quad II and cos β = 3/4 and lies in quad I
x = -12
y = 5
r = 13
x = 3
y = √7 (suppose to be square root of 7)
r = 4
tan (α-β) = (tan α - tan ß) / 1 + tan (α) tan (ß)
tan (α-β) = ((-5/12) - (√7/4)) / 1 + (-5/12)(√7/4)
tan (α-β) = ((-5/12) - (3√7/12)) / (48/48) + ((-5√7)/48)
tan (α-β) = ((-5-3√7)/12)((48)/(48-5√7)
tan (α-β) = -1/4 (is what I eventually came up with but it's wrong and I am unsure of what I'm doing wrong...)
I've also added a screen shot of the actual equation just in case the above is too confusing...
Thanks for replying back ~ but I have 2 additional questions.
1. before we rationalize, how did you get from the tan (α-β) equation to the (-15-12√7)/(36-5√7) ~ you multiplied it the whole equation by 36/36 right?
2. then when you rationalize the equation with 36+5√7 ~ this might be an algebra question, and I've probably forgotten, but why can't you just multiply by 5√7 instead of what you have?
Sorry but now Im confused ~ bc I thought that you would have to rationalize the denominator if there is a square root there... b/c if I didn't have to rationalize it in the first place, why even bother multiplying it by the 5√7 in the first place... or even the 36+5√7...
I'm sorry, but I think your causing me more confusion on this question Sorban rationalized the answer and I know I didn't post to rationalize the answer, but the attachment I posted showed that it needed to be rationalized. I am prob the one that caused the confusion on this question, but I tried to minimize it with posting the attachment. I think I will just PM Sorban for more information, thanks for your help anyway biffboy. I do appreciate it.