document.write( "Question 66295: The \"x's\" are actually thetas but for simplicity sake I am going to use \"x\" instead\r
\n" ); document.write( "\n" ); document.write( "tan^2x - sin^2x - tan ^2xsin^2x\r
\n" ); document.write( "\n" ); document.write( "I tried to replace the tan^2x's with (sin^2x)/(cos^2x) but I didn't know where to go from there.. or if that even works.\r
\n" ); document.write( "\n" ); document.write( "thanks alot.
\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #47013 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
tan^2x - sin^2x - tan ^2xsin^2x
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\n" ); document.write( "Then converting the tan^2 to sin^2/cos^2 you get:\r
\n" ); document.write( "\n" ); document.write( "(sin^2/cos^2)-sin^2 - (sin^2/cos^2)(sin^2)\r
\n" ); document.write( "\n" ); document.write( "Factor out the sin^2 to get:
\n" ); document.write( "(sin^2)[1/cos^2 - 1 -(sin^2/cos)^2]
\n" ); document.write( "Rewrite the 2nd factor with LCD=cos^2 to get:
\n" ); document.write( "= (sin^2)[1-cos^2-sin^2]/cos^2
\n" ); document.write( "= (sin^2/cos^2)[1-(cos^2+sin^2)]
\n" ); document.write( "But cos^2+sin^2=1, SO you get,
\n" ); document.write( "=(sin^2/cos^2)[1-1]
\n" ); document.write( "=0
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H. \n" ); document.write( "
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