document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=602403>Question 602403</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do you prove that this is an identity?\r<BR>\n" );
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document.write( "sinX/cosX + cosX/sinX = secX cscX\r<BR>\n" );
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document.write( "I've been thinking that in the end both sides should somehow end up with 1/(cosX)(sinX) but I really don't know how to do that since the left side has addition and not multiplication so maybe I'm wrong? IDK I'm so confused.\r<BR>\n" );
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document.write( "Thank you to whoever answers this! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #380235 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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=380235\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> (sin(x))/(cos(x))+(cos(x))/(sin(x)) = sec(x)csc(x)\r<BR>\n" );
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document.write( "(sin(x)sin(x))/(sin(x)cos(x))+(cos(x))/(sin(x)) = sec(x)csc(x)\r<BR>\n" );
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document.write( "(sin^2(x))/(sin(x)cos(x))+(cos(x))/(sin(x)) = sec(x)csc(x)\r<BR>\n" );
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document.write( "(sin^2(x))/(sin(x)cos(x))+(cos(x)cos(x))/(sin(x)cos(x)) = sec(x)csc(x)\r<BR>\n" );
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document.write( "(sin^2(x))/(sin(x)cos(x))+(cos^2(x))/(sin(x)cos(x)) = sec(x)csc(x)\r<BR>\n" );
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document.write( "(sin^2(x)+cos^2(x))/(sin(x)cos(x)) = sec(x)csc(x)\r<BR>\n" );
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document.write( "(1)/(sin(x)cos(x)) = sec(x)csc(x)\r<BR>\n" );
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document.write( "(1/sin(x))*(1/cos(x))= sec(x)csc(x)\r<BR>\n" );
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document.write( "csc(x)*sec(x)= sec(x)csc(x)\r<BR>\n" );
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document.write( "sec(x)csc(x)= sec(x)csc(x)\r<BR>\n" );
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document.write( "This verifies the identity. </SPAN>\n" );
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