(3/sin²40-1/cos²40)/(2sin10)求值,帮个忙,

问题描述:

(3/sin²40-1/cos²40)/(2sin10)求值,帮个忙,

先通分,在整理降幂,再用倍角公式等化简,从而得出答案 16

16

(3/sin²40-1/cos²40)/(2sin10)
=[(3cos²40-sin²40)/(sin²40cos²40)]/(2sin10)
=4[((√3cos40+sin40)(√3cos40-sin40)/sin²80]/(2sin10)
=[16sin(60+40)sin(60-40)/cos²10]/(2sin10)
=(16sin100sin20)/(2sin10cos²10)
=[(16cos10*2cos10sin10)/(2sin10cos²10)
=(32sin10cos²10)/(2sin10cos²10)
=16