SOLUTION: Pleease help me solve the identity!! {{{(1-cotx)/cscx + (cscx)/cotx = (1+cotx)/(cscxcotx)}}} Thanks so much in advance and please reply as soon as possible

Algebra ->  Trigonometry-basics -> SOLUTION: Pleease help me solve the identity!! {{{(1-cotx)/cscx + (cscx)/cotx = (1+cotx)/(cscxcotx)}}} Thanks so much in advance and please reply as soon as possible      Log On


   

Question 773689: Pleease help me solve the identity!!
%281-cotx%29%2Fcscx+%2B+%28cscx%29%2Fcotx+=+%281%2Bcotx%29%2F%28cscxcotx%29

Thanks so much in advance and please reply as soon as possible
Answer by trx1150(14) About Me  (Show Source):
You can put this solution on YOUR website!
When solving an identity, always start with the side that looks more complicated.
To me, I think that the left side looks more complicated. In the following work, I will only be manipulating the left side and leaving the right side out for now.

%281-cotx%29%2Fcscx+%2B+%28cscx%29%2Fcotx The denominator of the right side is cscxcotx so we must try and get the denominators to match up. To get a common denominator, we can multiply the left fraction by cotx and the right fraction by cscx.

%281-cotx%29%28cotx%29%2Fcotxcscx+%2B+%28cscx%29%28cscx%29%2Fcotxcscx
From here, let's simplify the expression by distributing where needed. Note that we can now combine the fractions since they have the same denominator but we will hold off on that temporarily.

%28cotx-cot%5E2x%29%2Fcotxcscx+%2B+%28csc%5E2x%29%2Fcotxcscx For the right fraction, let's rewrite csc%5E2x as 1%2Bcot%5E2x by using the pythagorean identity csc%5E2x=1%2Bcot%5E2x.

We know have %28cotx-cot%5E2x%29%2Fcotxcscx+%2B+%281%2Bcot%5E2x%29%2Fcotxcscx
Let's now combine the fractions.

%28cotx-cot%5E2x%2B1%2Bcot%5E2x%29%2Fcotxcscx And then finally combine like terms.

%281%2Bcotx%29%2Fcotxcscx+=+%281%2Bcotx%29%2Fcotxcscx
Let me know if you have any other questions.