document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=764203>Question 764203</A>: <I><SPAN STYLE=\"word-break: break-all;\"> if cos(theta) =1/2 find value of 2 sec(theta) divided by 1 plus tan square (theta) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #465278 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=465278\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> if cos(theta) =1/2 find value of 2 sec(theta) divided by 1 plus tan square (theta)<BR>\n" );
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document.write( "If cos(theta) = 1/2, x = 1 and r = 2<BR>\n" );
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document.write( "So y = sqrt[2^2-1^2] = sqrt(3)<BR>\n" );
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document.write( "Therefore, sec(theta) = 2 <BR>\n" );
document.write( "and tan(theta) = sin/cos = (sqrt(3))/(1/2) = 2sqrt(3)<BR>\n" );
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document.write( "Your Problem:<BR>\n" );
document.write( "2*sec(theta)/[1 + tan^2(theta)]<BR>\n" );
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document.write( "= 2(2)/(1 + (2sqrt(3))^2)<BR>\n" );
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document.write( "= 4/[1+12]<BR>\n" );
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document.write( "= 4/13<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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document.write( " </SPAN>\n" );
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