求证1/(sinˆ2xcosˆ2x)—2=2（3+cos4x）/(1—cos4x)

求证1/(sinˆ2xcosˆ2x)—2=2（3+cos4x）/(1—cos4x)
求证1/(sinˆ2xcosˆ2x)—2=2（3+cos4x）/(1—cos4x)

求证1/(sinˆ2xcosˆ2x)—2=2（3+cos4x）/(1—cos4x)
证明：
右边=2（3+cos4x）/(1—cos4x)
=(6+2cos4x)/(1—cos4x)
=(6+2cos4x)/[1-(1-2sin2x^2)]
=(6+2cos4x)/(1-1+2sin2x^2)
=(6+2cos4x)/2sin2x^2
=（3+cos4x）/sin2x^2
=(3+1-2sin2x^2)/sin2x^2
=(4-2sin2x^2)/sin2x^2
=4/sin2x^2-2sin2x^2/sin2x^2
=4/sin2x^2-2
=4/(2sinx*cosx)^2-2
=4/4sinˆ2x*cosˆ2x-2
=1/(sinˆ2xcosˆ2x)-2=左边
所以原式得证!