SOLUTION: I had a test today, and could not crack this problem I really would appreciate the help so I can understand it later on. Prove: cotx*cosx=cotx-cosx first i found a commo

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Question 187494: I had a test today, and could not crack this problem I really would appreciate the help so I can understand it later on.

Prove:
cotx*cosx=cotx-cosx

first i found a common denominator, but after that i was stuck.
Found 2 solutions by Alan3354, jim_thompson5910:
Answer by Alan3354(21605) About Me  (Show Source):
You can put this solution on YOUR website!
Prove:
cotx*cosx=cotx-cosx
I don't think it is equal.
cos/tan = cot - cos
cos = 1 - cos*tan
cos = 1 - sin That is not true.

Answer by jim_thompson5910(21685) About Me  (Show Source):
You can put this solution on YOUR website!
I'm assuming that you want to prove that cot%28x%29%2Acos%28x%29=cot%28x%29-cos%28x%29 is an identity right?


There's a problem here, the equation:


cot%28x%29%2Acos%28x%29=cot%28x%29-cos%28x%29


is NOT an identity (graph the two sides if you aren't sure)

--------------


For example, let x=1


cot%28x%29%2Acos%28x%29=cot%28x%29-cos%28x%29 Start with the given equation.


cot%281%29%2Acos%281%29=cot%281%29-cos%281%29 Plug in x=1


%280.642%29%2A%280.540%29=0.642-0.540 Evaluate the given trig functions (note: I'm working in radian mode).


0.34668=0.102 Simplify. Clearly the two sides are NOT equal.


Since 0.34668%3C%3E0.102, this means that the equation is NOT true for ALL values of "x". So the equation is NOT an identity.



So either there's a typo or you were supposed to disprove it being an identity. My guess would be that there's a typo in there somewhere.


Note: if you change the "cot" to "csc" and "cos" to "sin" on the right, you'll get cot%28x%29%2Acos%28x%29=csc%28x%29-sin%28x%29 which is an identity