document.write( "Question 368957: Ok I'm having trouble proving these harder identities.
\n" ); document.write( "The homework says prove that each of the following identities is true
\n" ); document.write( "(2sec^2)x-(2sec^2)x(sin^2)x-(sin^2)x-(cos^2)x=1
\n" ); document.write( "It's w/o the parenthesis in the homework, I just didn't know how to seperate the squares for the x's.
\n" ); document.write( "I tried multiplying by (2cos^2)x to get rid of the sec's but I'm getting nowhere with it. Here's what I did even though I know it's wrong:
\n" ); document.write( "*-1 then (2cos^2)x
\n" ); document.write( "=-(2cos^4)x+(4cos^2)x(sin^2)x=0
\n" ); document.write( "/(2cos^2)x
\n" ); document.write( "=(-cos^2)x+(2cos^2)x(sin^2)x=0
\n" ); document.write( "Then I don't see anywhere to go!
\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #262900 by stanbon(75887)\"\" \"About 
You can put this solution on YOUR website!
(2sec^2)x-(2sec^2)x(sin^2)x-(sin^2)x-(cos^2)x=1
\n" ); document.write( "-------------------------------------------------
\n" ); document.write( "2sec^2 - 2sec^2*sin^2 -[sin^2+cos^2] = 1
\n" ); document.write( "----
\n" ); document.write( "2sec^2[1-sin^2] - 1 = 1
\n" ); document.write( "----
\n" ); document.write( "2sec^2[cos^2] = 2
\n" ); document.write( "---
\n" ); document.write( "2(1) = 2
\n" ); document.write( "2 = 2
\n" ); document.write( "==========
\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H. \n" ); document.write( "
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