document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=681990>Question 681990</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Prove that (1-tan^2 x) / (1+tan^2 x) = cos 2 x \r<BR>\n" );
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document.write( "For each step of the proof I also need to know what rule this applies to (algebra, reciprocal, quotient, Pythagorean, double angle, or sum and differences). Please state the rule after each step\r<BR>\n" );
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document.write( "Thank you so so much for your help! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #422930 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=422930\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Prove that (1-tan^2 x) / (1+tan^2 x) = cos 2 x <BR>\n" );
document.write( "-----<BR>\n" );
document.write( "Note: Since sin^2 + cos^2 = 1<BR>\n" );
document.write( "Dividing by cos^2 you get: tan^2 + 1 = sec^2<BR>\n" );
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document.write( "Using that Pythagorean relation you get:<BR>\n" );
document.write( "(1-tan^2)/(sec^2) = cos(2x)<BR>\n" );
document.write( "------<BR>\n" );
document.write( "(1 - (sin^2/cos^2)/sec^2 = cos(2x)<BR>\n" );
document.write( "(cos^2-sin^2) = cos(2x)<BR>\n" );
document.write( "cos(2x) = cos(2x) </SPAN>\n" );
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