SOLUTION: verify the identity algebraically (1-2sin^2 x)/(sinxcosx) = cotx-tanx

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Question 596215: verify the identity algebraically
(1-2sin^2 x)/(sinxcosx) = cotx-tanx

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
%281-2sin%5E2%28x%29%29%2F%28sin%28x%29cos%28x%29%29+=+cot%28x%29-tan%28x%29
There are a number of things to consider when trying to figure out these identities:
  • Match arguments -- Use argument-changing Trig properties (2x, 1/2x, A+B, A-B) to change arguments on the left to match those on the right
  • Match the number of terms -- Use algebra and/or Trig properties to get the same number of terms on the left side as there are on the right.
  • Match functions -- Use Trig properties to change functions on the left to match those on the right.
  • If none of the above look like they are going to work, try using Trig properties to change sec, csc, tan or cot on the left into sin and/or cos.
Unfortunately, I cannot give a recipe of what to do. You just have to know your properties and algebra well enough to see which of the above looks promising. (If nothing looks promising, change everything into sin's and/or cos's and look again.)

The arguments match. So we are looking to match the number of terms (1 on the left and 2 on the right) and to turn the sin's and cos's on the left into a cot and a tan. If "un-subtract" the fraction to get two terms we get:

But the path from here to the end is not clear to me. So I returned to:
%281-2sin%5E2%28x%29%29%2F%28sin%28x%29cos%28x%29%29+=+cot%28x%29-tan%28x%29
and started looking for something else. I noticed that the numerator on the left is one of the forms of the cos(2x) formula. We do not want to introduce a new argument, 2x. But we can replace the numerator with one of the other forms of the cos(2x) formula:
  • cos%5E2%28x%29-sin%5E2%28x%29
  • 2cos%5E2%28x%29-1
  • 1-2sin%5E2%28x%29
Let's replace 1-2sin%5E2%28x%29 with cos%5E2%28x%29-sin%5E2%28x%29:
%28cos%5E2%28x%29-sin%5E2%28x%29%29%2F%28sin%28x%29cos%28x%29%29+=+cot%28x%29-tan%28x%29

Now let's see what happens if we "un-subtract":


In the first fraction cos(x) will cancel and in the second fraction sin(x) will cancel:
cos%28x%29%2Fsin%28x%29-sin%28x%29%2Fcos%28x%29+=+cot%28x%29-tan%28x%29
which is what we want:
cot%28x%29-tan%28x%29+=+cot%28x%29-tan%28x%29