document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=234512>Question 234512</A>: <I><SPAN STYLE=\"word-break: break-all;\"> plese help prove: (cot(x))/(csc(x)+1)=(csc(x)-1)/(cot(x)) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #172913 by </B><A HREF=/tutors/aboutme.mpl?userid=nyc_function><B>nyc_function(2741)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=nyc_function><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=nyc_function><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=172913\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The best way to handle trig identities is to convert everything to sine and cosine.\r<BR>\n" );
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document.write( "It is always best to start with the hardest side but in this case, it makes no difference because both sides of the equation are complicated.\r<BR>\n" );
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document.write( "Here is the conversion:\r<BR>\n" );
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document.write( "cotx = cosx/sinx\r<BR>\n" );
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document.write( "cscx = 1/sinx\r<BR>\n" );
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document.write( "I will set it up and you can finish.\r<BR>\n" );
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document.write( "[(cosx/sinx)]/[(1/sinx)  + 1)] = [(1/sinx) - 1)]/(cosx/sinx)\r<BR>\n" );
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document.write( "This ugly thing is just a complex fraction.  \r<BR>\n" );
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document.write( "Can you take it from here?\r<BR>\n" );
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document.write( " </SPAN>\n" );
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