document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=682945>Question 682945</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I need help sloving this identity, please show the steps becasue I subed in 2 for B and both sides were the same so I know it is an identity but I do not how to prove it without subing in a number\r<BR>\n" );
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document.write( "(1-cosB)/(sinB)=(sinB)/(1+cosB)\r<BR>\n" );
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document.write( "Thanks :) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #423400 by </B><A HREF=/tutors/aboutme.mpl?userid=lwsshak3><B>lwsshak3(11628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=lwsshak3><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=423400\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> prove following identity:<BR>\n" );
document.write( "(1-cosB)/(sinB)=(sinB)/(1+cosB)<BR>\n" );
document.write( "stare with left side<BR>\n" );
document.write( "(1-cosB)/(sinB)<BR>\n" );
document.write( "multiply both top and bottom by 1+cosB, making the denominator a difference of 2 squares<BR>\n" );
document.write( "(1-cosB)/(sinB)*1+cosB/1+cosB<BR>\n" );
document.write( "=(1-cos^2B)/sinB(1+cosB)<BR>\n" );
document.write( "=sin^2B/(sinB(1+cosB))<BR>\n" );
document.write( "=sinB/1+cosB<BR>\n" );
document.write( "verified: left side=right side<BR>\n" );
document.write( "  </SPAN>\n" );
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