document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1045319>Question 1045319</A>: <I><SPAN STYLE=\"word-break: break-all;\"> secA+cosA is not equal to 3/2 prove\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #660719 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=660719\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> sec A and cos A are reciprocals.<BR>\n" );
document.write( "x+(1/x)=3/2.  See if this is possible.<BR>\n" );
document.write( "common denominator (x^2+1)/x=3/2<BR>\n" );
document.write( "cross multiply and 3x=2x^2+2<BR>\n" );
document.write( "2x^2-3x+2=0<BR>\n" );
document.write( "the discriminant b^2-4ac is 9-16, which is negative, so there are no real roots.<BR>\n" );
document.write( "Therefore a number plus its reciprocal cannot equal 3/2.<BR>\n" );
document.write( "sec A is a number and cos A is its reciprocal, and they cannot equal 3/2.<BR>\n" );
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