document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=839790>Question 839790</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If tan 0=2 with 0 in quadrant 111 , find sec 0? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #505852 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=505852\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  tan(t)=2 with t in quadrant 111 , find sec(t)0?<BR>\n" );
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document.write( "Note: Both x and y are negative in QIII<BR>\n" );
document.write( "Since tan = y/x, y = -2 and x = -1<BR>\n" );
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document.write( "r = sqrt(2^2+1^2] = sqrt(5)<BR>\n" );
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document.write( "Your Problem::<BR>\n" );
document.write( "sec(t) = r/x = sqrt(5)/-1 = -sqrt(5)<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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