document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=135374>Question 135374</A>: <I><SPAN STYLE=\"word-break: break-all;\"> (2sinxcosx + cosx - 2sinx - 1)/(-sin(squared)x) = (2 sinx + 1)/(cosx+1)<BR>\n" );
document.write( "Please prove one side to equal the other (verify identity). Thank you. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #99213 by </B><A HREF=/tutors/aboutme.mpl?userid=kev82><B>kev82(151)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=kev82><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=99213\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> It's not exactly an identity as for x=0 the LHS is not evaluable but the RHS is. Similar for x=pi.\r<BR>\n" );
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document.write( "Let S=sin, C=cosine then the first thing I will do is multiply the whole thing by SS(C+1). This gives\r<BR>\n" );
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document.write( "-(C+1)(2SC+C-2S-1)=(2S+1)SS\r<BR>\n" );
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document.write( "expanding all the brackets gives\r<BR>\n" );
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document.write( "-(2SCC + CC - 2SC -C +2SC + C - 2S - 1) = 2SSS + SS\r<BR>\n" );
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document.write( "With some cancellations\r<BR>\n" );
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document.write( "-(2SCC + CC - 2S - 1) = 2SSS + SS\r<BR>\n" );
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document.write( "Now CC + SS =1, so CC = 1-SS, substituting this in gives\r<BR>\n" );
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document.write( "-(2S - 2SSS + 1 - SS - 2S - 1) = 2SSS + SS\r<BR>\n" );
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document.write( "From here it is just trivial cancellation to see that both are the same. </SPAN>\n" );
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