Question 737700
(1-sin^4x) is the difference of two squares.  It equals (1-sin^2x)(1+sin^2x)<P>
sec^2x = 1/cos^2x and tan^2x = sin^2x/cos^2x<P>
sin^2x + cos^2x =1 and 1-Sin^2x = cos^2x
(1-sin^4x)/(sec^2x+tan^2x) = (1-sin^2x)(1+sin^2x)/(1/cos^2x + sin^2x/cos^2x) = (1-sin^2x)(1+sin^2x)/ (1+sin^2x)/cos^2x=<P>
(1-sin^2x)(1+sin^2x) * cos^2x/(1+sin^2x):  The 1+sin^2x cancel.<P>
(1-sin^2x)* cos^2x = cos^2x * cos^2x = cos^4x
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