If sin⁡θ,tan⁡θ,cos⁡θ are in G.P. then 4sin2⁡θ−3sin4⁡θ+sin6⁡θ=

Given that tan2⁡θ=sin⁡θcos⁡θ∴ sin⁡θ=cos3⁡θAlso, given expression can be rewritten as sin6⁡θ−3sin4⁡θ+3sin2⁡θ−1+1+sin2⁡θ=sin2⁡θ−13+1+sin2⁡θ=−1−sin2⁡θ3+1+sin2⁡θ=−cos6⁡θ+1+sin2⁡θ=−sin2⁡θ+1+sin2⁡θ=1