已知θ为锐角,则logsinθ(1+cot^2θ)=

1+cot^2θ=csc^2θ=1/sin^2θ=(sinθ)^(-2)

所以原式=-2

1+cot^2θ等于1/sinθ
所以答案是-1

1+(cotθ)^2=(cscθ)^2=1/(sinθ)^2

则原式=logsinθ[1/(sinθ)^2]=-2

1+cot^2A=(cos^2A+sin^2A)/sin^2A=(sinA)^-2

则logsinθ(1+cot^2θ)=-2

1+cot²θ
=1+(1/tan²θ)
=1+(cos²θ/sin²θ)
=(sin²θ/sin²θ)+(cos²θ/sin²θ)
=(sin²θ+cos²θ)/sin²θ
=1/sin²θ
=(sinθ)^(-2)
logsinθ(1+cot²θ)
=logsinθ(sinθ)^(-2)
=(-2)logsinθsinθ
=(-2)*1
=-2