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Trigonometry

tanU-1/tanU+1=1-cotU/1+cotU

So, to correct your question, no, you don't show that "the answer is 1", but you do show that IF the original statement is true, THEN 1=1 (or any other identity statement), --> *and vice versa* <-- , that is, IF 1=1 (which you know it does), THEN the original statement was true. If you use correct math, and are careful not to do anything that multiplies or divides both sides of an equation by zero, then both of the IF/THEN statements above will be equally true. That is, if you can break down the original statement into a true identity, then the original statement is proved true also.

Arthur D. | Effective Mathematics TutorEffective Mathematics Tutor
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(tanU-1)/(tanU+1)=(1-cotU)/(1+cotU)
tanU=1/cotU
[1/(cotU)-1]/[1/(cotU)+1]
[(1-cotU)/(cotU)]/[(1+cotU)/(cotU)]
(1-cotU)/(1+cotU)
dividing by a number is the same as multiplying by its reciprocal so cotU and cotU cancel

Stanton D. | Tutor to Pique Your Sciences InterestTutor to Pique Your Sciences Interest
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Kayla,
When you submit a problem, you'd be doing everyone a favor if you would parenthesize your expressions, since you can't write them as fractions.
I'm guessing that you mean (tanU-1)/(tanU +1) = (1-cotU)/(1+cotU) ??
if so, rephrase the relationships in terms of x and y lengths of a right triangle:
((y/x)-(x/x))/((y/x)+(x/x)) = ((y/y)-(x/y))/((y/y)+(x/y))
this reduces to
(y-x)/(y+x) = (y-x)/(y+x)
or 1=1 {{if neither of x,y  = 0 and x not= -y; you always have to go back in a problem like this and make sure that nothing (i.e. no single variable or expression) that was a denominator was =0, because that would be division by zero, which isn't allowed!}}