Kayla M.

asked • 03/10/14

Trigonometry

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

Stanton D.

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.
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03/12/14

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Stanton D. answered • 03/10/14

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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!}}
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Kayla M.

So the answer to that equation would be 1? 
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03/10/14

Arthur D.

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You have to show that the left hand side equals the right hand side using trig identities. Look at my solution.
Start with the left hand side and use trig identities to change the left hand side into the right hand side.
 
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03/10/14

Stanton D.

Actually, Arthur and I have demonstrated two approaches to the proof which use different appearances but both work (incidentally, both these types of methods also apply for proving other mathematical assertions). To prove the equality that was the original problem, you can either manipulate only one side of the equation, using standard algebra (namely, that cotU = 1/tanU) to turn it into the other side -- that's Arthur's approach -- or, you can work with both sides of the equation, to turn the proposed equality into an identity statement (such as, 1=1). Either way works, and they are, at heart, identical math (Arthur just expresses his as trig functions, and I express mine as the geometric equivalent of the trig functions). This is not like having different proofs; it's just a matter of convenience which way works better for you.
Incidentally, this is somewhat like how the transitive property of equality works: if A=B, and B=C, then A=C, right? Or, to use another analogy, it's like converting a measurement in one system of units into an equivalence in another system of units -- you may have to multiply by conversion factors to get the first value into a form that you know how to step across into the second system, then work it up to the desired units. You could do all that; but, if you were asked if two given values in the two different systems were equivalent, you could just break them both down into the "step-across" equivalents, i.e. work the conversion from both ends, meeting at the middle, rather than working it all the way from one end to the other. That's kind of like what reducing the proposed equality to an identity statement does. OK?
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03/12/14

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