If tan⁡π−2α4⋅tan⁡π−2β4⋅tan⁡π−2γ4=1, then the value of 212(sin⁡α⋅sin⁡β⋅sin⁡γ+sin⁡α+sin⁡β+sin⁡γ) is

Given that tan⁡π−2α4tan⁡π−2β4tan⁡π−2γ4=1⇒tan2⁡π4−α2tan2⁡π4−β2tan2⁡π4−γ2=1⇒1−tan⁡α21+tan⁡α221−tan⁡β21+tan⁡β221−tan⁡γ21+tan⁡γ22=1⇒1−sin⁡α1+sin⁡α⋅1−sin⁡β1+sin⁡β1−sin⁡γ1+sin⁡γ=1⇒sin⁡α+sin⁡β+sin⁡γ+sin⁡α⋅sin⁡β⋅sin⁡γ=0∴212(sin⁡αsin⁡βsin⁡γ+sin⁡α+sin⁡β+sin⁡γ)=0