document.write( "Question 249656: Verify the identity
\n" ); document.write( "sin(x)/(1+cos(x))+(1+cos(x))/sin(x)=2csc(x) \n" ); document.write( "
Algebra.Com's Answer #181809 by jsmallt9(3758)\"\" \"About 
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\"sin%28x%29%2F%281%2Bcos%28x%29%29%2B%281%2Bcos%28x%29%29%2Fsin%28x%29=2csc%28x%29\"

\n" ); document.write( "Since the right side has 1 term and the left side has two terms, we will start by adding the terms on the left. Of course to add fractions we must have common denominators:
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\n" ); document.write( "which simplifies:
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\n" ); document.write( "Now we can add:
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\n" ); document.write( "Since \"sin%28x%29%5E2+%2B+cos%28x%29%5E2+=+1\" we get:
\n" ); document.write( "\"%281+%2B+1%2B2cos%28x%29%29%2F%28%281%2Bcos%28x%29%29%28sin%28x%29%29%29+=+2csc%28x%29\"
\n" ); document.write( "\"%282+%2B+2cos%28x%29%29%2F%28%281%2Bcos%28x%29%29%28sin%28x%29%29%29+=+2csc%28x%29\"
\n" ); document.write( "We now have a simplified, one-term expression on the left. Since the right side has a factor of 2, we'll factor out a 2 on the left:
\n" ); document.write( "\"%282%281+%2B+cos%28x%29%29%29%2F%28%281%2Bcos%28x%29%29%28sin%28x%29%29%29+=+2csc%28x%29\"
\n" ); document.write( "As you can see, the (1 + cos(x))'s cancel leaving:
\n" ); document.write( "\"2%2Fsin%28x%29+=+2csc%28x%29\"
\n" ); document.write( "or
\n" ); document.write( "\"2%281%2Fsin%28x%29%29+=+2csc%28x%29\"
\n" ); document.write( "Since 1/sin(x) = csc(x) we have:
\n" ); document.write( "\"2csc%28x%29+=+2csc%28x%29\"
\n" ); document.write( "And we are finished.
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