# Trig ((tanx - sinx)/sin^3x) = (1/(cosx - cos^2x))

1. Apr 10, 2014

### Attis

1. The problem statement, all variables and given/known data

show that ((tanx - sinx)/sin^3x) = (1/(cosx - cos^2x))

2. Relevant equations
sin^2 = 1-cos^2v
cos^2 = 1-sin^2v

3. The attempt at a solution
On the left hand side: I plugged in sinx/cosx for tanx. I then divided both the numerator and denominator by sinx to get ((1/cosx - 1)/ sin^2x).
After this point I got stuck. I tried plugging in 1-cos^2x for sin^2x and simplifying the fraction, but it didn´t work. Do you guys have any tips on how to progress?

2. Apr 10, 2014

### Staff: Mentor

are you sure the righthand side is 1/(cosx - cos^2x). Im thinking it should be 1/(cosx + cos^2).

I started by assuming its true and then doing the crossproduct trick ie a/b = c/d ==> ad = bc and then used algebra to reduce it down and from there you can see how to derive it.

3. Apr 10, 2014

### az_lender

Both of you are writing nonsense! "cos^2" or "sin^2" without an angle argument such as "x" is absolutely meaningless. (Both in jedishrfu's answer and in attis's "relevant equations".

Apparently what's meant is
[tan(x) - sin(x)]/sin3(x) = 1/[cos(x) - cos2(x)]

If you divide a factor of sin(x) out of both the numerator and the denominator on the left, you have
[sec(x) - 1] / sin2(x) = 1/[cos(x) - cos2(x)]
If you then multiply both sides by cos(x) sin2(x), you have
1 - cos(x) = sin2(x) / [1 - cos(x)]

And now I see jedishrfu's point, that this is not going to come out to an identity unless the two remaining binomials include one sum and one difference.

BTW, shouldn't a trig question be in "precalculus" homework?

4. Apr 10, 2014

### Staff: Mentor

Thanks for pointing out my mistake. I think in both of our cases it was a simple typo to leave out the x argument and not nonsense. Please be kind when you respond and try to understand the intent of the poster. Sometimes autocorrect will interfere with your typing and cause these kinds of errors (but not in this case i simply missed typing the x).

Also, we try to help here without giving away too much of the answer until we see some effort on a part of the original poster. It took me awhile to see the wisdom of this approach.

5. Apr 11, 2014

### Attis

You were right about my typo. It should´ve been 1/(cosx + cos^2x). Sorry! I used the crossproduct trick and algebra... and it worked! Super. I don´t know why the other guy is calling it nonsense.
Thanks! Have a nice weekend.