Question

log(tanθ+cotθ)sinθ=-1/4,θ∈(0,π／2),则log(tanθ)sinθ=?？

Answer 1

因为(tanθ+cotθ)=1/(sinθ*cosθ),
所以,由log(tanθ+cotθ)sinθ=-1/4,得：
log(sinθ) (tanθ+cotθ)=-4,
log(sinθ) [1/(sinθ*cosθ)]=-4,
log(sinθ) (sinθ*cosθ)=4,
1+log(sinθ) (cosθ)=4,
log(sinθ) (cosθ)=3,
log(sinθ) (tanθ)=1-log(sinθ) (cosθ)=-2,
log(tanθ)sinθ=-1/2.,7,log(tanθ+cotθ)sinθ=(lgsinθ)/[lg(sinθ/cosθ+cosθ/sinθ)]
=(lgsinθ)/lg[1/sinθcosθ]=-(lgsinθ)/lg[sinθcosθ]=-(lgsinθ)/[lgsinθ+logcosθ]=-1/4
可得3logsinθ=lgcosθ
log(tanθ)sinθ=lgsinθ/[logsinθ-lgcosθ]=-1/2,1,log(tanθ+cotθ)sinθ=-1/4,θ∈(0,π／2),则log(tanθ)sinθ=?
需要过程.谢谢.