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

Triangle α

x = -12

y = 5

r = 13

Triangle β

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...