Given the root 2 of (1 + TaNx) / (1-tanx) = 3 + 2 times, find the values of sinxcosx and SiNx (sinx-3cosx)

Given the root 2 of (1 + TaNx) / (1-tanx) = 3 + 2 times, find the values of sinxcosx and SiNx (sinx-3cosx)

tanx=√2/2,
So SiNx = √ 3 / 3, cosx = √ 6 / 3
So sinxcosx = √ 2 / 3
sinx(sinx-3cosx)=1/3-√2