how to find the 6 trigo values of -5,12?
-5, 12......and r will = sqrt [ (-5)^2 + 12^2 ] = sqrt[169] = 13
I'm assuming that x = -5 and y = 12
So...
sin = y/r = 12/13
csc = r/y = 13/12
cos = x/r = -5/13
sec = r/x = -13/5
tan = y/x =-12/5
cotangent = x/y = -5/12
