Question

(1+tan12)(1+tan22)(1+tan23)(1+tan33)怎么算

Answer 1

设(1 + tan12)(1 + tan22)(1 + tan23)(1 + tan33)= A/B

B = cos21*cos22*cos23*cos24

A = (sin12 + cos12)(sin22 + cos22)(sin23 + cos23)(sin33 + cos33)

= 4*sin(12 + 45)*sin(22 + 45)*sin(23 + 45)*sin(33 + 45)

= 4*sin57*sin67*sin68*sin78

= 4*cos33*cos23*cos22*cos12

A/B = 4

所以 (1 + tan12)(1 + tan22)(1 + tan23)(1 + tan33)=4