L.H.S
1+cosθsinθ+sinθ1+cosθ
=(1+cosθ)2+sin2θsinθ(1+cosθ)
=1+cos2θ+2cosθ+sin2θsinθ(1+cosθ)
=1+cos2θ+sin2θ+2cosθsinθ(1+cosθ)(Since:sin2x+cos2x=1)
=1+1+2cosθsinθ(1+cosθ)
=2+2cosθsinθ(1+cosθ)
=2(1+cosθ)sinθ(1+cosθ)
=2sinθ
=2cscθ
Hence, proved.