求证1/sin2θ + 1/tan2θ +1/sinθ =1/tanθ/2

求证1/sin2θ + 1/tan2θ +1/sinθ =1/tanθ/2
求证1/sin2θ + 1/tan2θ +1/sinθ =1/tanθ/2

求证1/sin2θ + 1/tan2θ +1/sinθ =1/tanθ/2
因为 1/sin2θ + 1/tan2θ =1/sin2θ +cos2θ/sin2θ
=(1+cos2θ)/sin2θ=2(cosθ)^2/(2sinθcosθ)
=cosθ/sinθ
所以1/sin2θ + 1/tan2θ +1/sinθ
=cosθ/sinθ+1/sinθ
=(1+cosθ)/sinθ
=[2cos(θ/2)^2]/[2sin(θ/2)cos(θ/2)]
=[cos(θ/2)]/[sin(θ/2)]
=1/tan(θ/2)