document.write( "Question 773689: Pleease help me solve the identity!!\r
\n" ); document.write( "\n" ); document.write( "\"%281-cotx%29%2Fcscx+%2B+%28cscx%29%2Fcotx+=+%281%2Bcotx%29%2F%28cscxcotx%29\"\r
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\n" ); document.write( "\n" ); document.write( "Thanks so much in advance and please reply as soon as possible
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Algebra.Com's Answer #471796 by trx1150(14)\"\" \"About 
You can put this solution on YOUR website!
When solving an identity, always start with the side that looks more complicated.
\n" ); document.write( "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.\r
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\n" ); document.write( "\n" ); document.write( "\"%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.\r
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\n" ); document.write( "\n" ); document.write( "\"%281-cotx%29%28cotx%29%2Fcotxcscx+%2B+%28cscx%29%28cscx%29%2Fcotxcscx\"
\n" ); document.write( "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.\r
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\n" ); document.write( "\n" ); document.write( "\"%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\".\r
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\n" ); document.write( "\n" ); document.write( "We know have \"%28cotx-cot%5E2x%29%2Fcotxcscx+%2B+%281%2Bcot%5E2x%29%2Fcotxcscx\"
\n" ); document.write( "Let's now combine the fractions.\r
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\n" ); document.write( "\n" ); document.write( "\"%28cotx-cot%5E2x%2B1%2Bcot%5E2x%29%2Fcotxcscx\" And then finally combine like terms.\r
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\n" ); document.write( "\n" ); document.write( "\"%281%2Bcotx%29%2Fcotxcscx+=+%281%2Bcotx%29%2Fcotxcscx\"\r
\n" ); document.write( "\n" ); document.write( "Let me know if you have any other questions.
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