document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=857812>Question 857812</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Could you please help me evaluate the following function: If the csc x = (4sqrt(3))/3 and the cos x < 0, what is the tan x<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 #516865 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=516865\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> evaluate the following function: If the csc t = (4sqrt(3))/3 and the cos t < 0, what is the tan t<BR>\n" );
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document.write( "Note:: Since csc is positive and cos is negative, t is in QII<BR>\n" );
document.write( "where x is negative and y = positive.<BR>\n" );
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document.write( "Since csc = r/y, r = 4sqrt(3) and y = 3<BR>\n" );
document.write( "Then x = -sqrt[(4sqrt(3))^2 + 3^2] = -sqrt[57]<BR>\n" );
document.write( "So:<BR>\n" );
document.write( "sin = 3/4sqrt(3)<BR>\n" );
document.write( "cos = -sqrt(57)/4sqrt(3) = -(1/4)sqrt(19)<BR>\n" );
document.write( "tan = y/x = -3/sqrt(57)\r<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
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