Question

已知(tanx)^2+(cotx)^2=m(m≥2),求(3+cos4x)/(1-cos4x)

Answer 1

sin²x/cos²x+cos²x/sin²x=m

[(sinx)^4+cosx)^4]/sin²xcos²x=m

[(sin²x+cosx²x)²-2sin²xcos²x]/sin²xcos²x=m

(1-2sin²xcos²x)/sin²xcos²x=m

(1/sin²xcos²x)-2=m

sin²xcos²x=1/(m+2)

4sin²xcos²x=4/(m+2)

sin²2x=4/(m+2)

cos4x=1-2sin²2x=1-8/(m+2)=(m-6)/(m+2)

所以原式=[3+(m-6)/(m+2)]/[1-(m-6)/(m+2)]

=m/2