document.write( "Question 370078: how do you prove that sinxCosx=cosx-cos^3x/sinx are egual?
\n" ); document.write( "also 2+2Cot^2= 2cotx secx cscx?
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Algebra.Com's Answer #263720 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
Let \"S=sin%28x%29\",\"C=cos%28x%29\" to reduce some of the clutter,
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\n" ); document.write( "\"S%5E2%2BC%5E2=1\"
\n" ); document.write( "\"S%5E2C%2BC%5E3=C\"
\n" ); document.write( "\"S%5E2C=C-C%5E3\"
\n" ); document.write( "\"SC=%28C-C%5E3%29%2FS\"
\n" ); document.write( "Now just go backwards.
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\n" ); document.write( "\"S%5E2%2BC%5E2=1\"
\n" ); document.write( "\"1%2BC%5E2%2FS%5E2=1%2FS%5E2\"
\n" ); document.write( "\"2%281%2BC%5E2%2FS%5E2%29=2%2FS%5E2\"
\n" ); document.write( "\"2%281%2BC%5E2%2FS%5E2%29=2%281%2FS%5E2%29%28C%2FC%29\"
\n" ); document.write( "\"2%281%2BC%5E2%2FS%5E2%29=2%28C%2FS%29%281%2FC%29%281%2FS%29\"
\n" ); document.write( "\"2%2B2%28C%5E2%2FS%5E2%29=2%28C%2FS%29%281%2FC%29%281%2FS%29\"
\n" ); document.write( "Again work backwards.
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