设函数y=x^-4px-2的图像过M(tanA,1) N(tanB,1)两点2cos2Acos2B+psin2(A+B)+2sin^(A-B)求的值

设函数y=x²-4px-2的图像过M(tanA,1) N(tanB,1)两点，
求2cos2Acos2B+psin2(A+B)+2sin²(A-B)的值

由已知--->tanA、tanB时二次方程x²-4px-3=0的两根
--->tanA+tanB=4p，tanAtanB=-3
--->tan(A+B)=(tanA+tanB)/(1-tanAtanB)=p
--->sin2(A+B)=2p/(1+p²)，cos2(A+B)=(1-p²)/(1+p²)

2cos2Acos2B+psin2(A+B)+2sin²(A-B)
= 2cos2Acos2B+psin2(A+B)-cos2(A-B)+1
= 2cos2Acos2B+psin2(A+B)-[cos2Acos2B+sin2Asin2B]+1
= [cos2Acos2B-sin2Asin2B]+psin2(A+B)+1
= cos2(A+B)+psin2(A+B)+1
= (1-p²)/(1+p²)+2p²/(1+p²)+1
= 2