document.write( "Question 1075296: Please help me prove this equation\r
\n" ); document.write( "\n" ); document.write( "((1-cos4x)/(1+cos4x)) + 1 = sec^2(2x) \n" ); document.write( "
Algebra.Com's Answer #689977 by Aaragorn(2)\"\" \"About 
You can put this solution on YOUR website!
We know that,
\n" ); document.write( "cos 2x = cos^2(x) - sin^2(x) = 2cos^2(x) - 1 = 1 - 2sin^2(x)\r
\n" ); document.write( "\n" ); document.write( "Therefore,
\n" ); document.write( "cos 2x = 2cos^2(x) - 1
\n" ); document.write( "=> 1 + cos 2x = 2cos^2(x)\r
\n" ); document.write( "\n" ); document.write( "Also,
\n" ); document.write( "cos 2x = 1 - 2sin^2(x)
\n" ); document.write( "=> 1 - cos 2x = 2sin^2(x)\r
\n" ); document.write( "\n" ); document.write( "Therefore,
\n" ); document.write( "1 - cos 4x = 1 - cos 2(2x) = 2sin^2(2x)
\n" ); document.write( "1 + cos 4x = 1 + cos 2(2x) = 2sin^2(2x)\r
\n" ); document.write( "\n" ); document.write( "Left had side of equation is..
\n" ); document.write( "(1 - cos 4x)/(1 + cos 4x) + 1
\n" ); document.write( "=> [ 2sin^2(2x) ] / [ 2sin^2(2x) ] + 1
\n" ); document.write( "=> tan^2(2x) + 1 \r
\n" ); document.write( "\n" ); document.write( "By formula, tan^2(x) + 1 = sec^2(x)\r
\n" ); document.write( "\n" ); document.write( "Therefore,
\n" ); document.write( "tan^2(2x) + 1 = sec^2(2x) = Right hand side\r
\n" ); document.write( "\n" ); document.write( "[HENCE PROVED]
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