document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=969502>Question 969502</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Prove. (cotx-cscx)(cosx+1)=sinx </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #592376 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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/cgi-bin/show-face.mpl?user=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=592376\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> cot x=cos x/sin x; csc x=1/sinx\r<BR>\n" );
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document.write( "Rewrite in terms of sin x and cos x\r<BR>\n" );
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document.write( "(cos x/sin x)-(1/sin x) (cos x +1)\r<BR>\n" );
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document.write( "There is a common denominator, sin x\r<BR>\n" );
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document.write( "[(cos x-1)/sin x] (cos x +1).  Now, multiply the first term by cos x  and by +1\r<BR>\n" );
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document.write( "{(cos^2 x-cos x)/sin x}  + (cos x- 1)/sin x\r<BR>\n" );
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document.write( "You have a common denominator of sin x\r<BR>\n" );
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document.write( "[cos^2 x- cos x + cos x -1]/sin x\r<BR>\n" );
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document.write( "In the numerator, the middle 2 terms disappear, and we have cos^2 x -1 left.  But that is sin^2 x<BR>\n" );
document.write( "because sin^2 x + cos ^2 x =1\r<BR>\n" );
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document.write( "we have sin^2 x/sin x .  That equals sin x, which is the other side of the equation.  \r<BR>\n" );
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document.write( " </SPAN>\n" );
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