document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=995590>Question 995590</A>: <I><SPAN STYLE=\"word-break: break-all;\"> given tan&#952; = &#8730;5, find sec&#952; and cot&#952; in quadrant III </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #614300 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=614300\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> given tan&#952; = &#8730;5, find sec&#952; and cot&#952; in quadrant III <BR>\n" );
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document.write( "Note: x and y are both negative in QIII<BR>\n" );
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document.write( "By definition, tan = y/x<BR>\n" );
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document.write( "So, y = -sqrt(5) and x = -1<BR>\n" );
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document.write( "Then, r = sqrt[(sqrt(5))^2+(1)^2] = sqrt[5+1] = sqrt(6)<BR>\n" );
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document.write( "Ans: sec(t) = r/x = -sqrt(6) and cot(t) = x/y = 1/sqrt(5) = (1/5)sqrt(5)<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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