If tan⁡(α+β−γ)tan⁡(α−β+γ)=tan⁡γtan⁡β,(β≠γ) then sin⁡2α+sin⁡2β+sin⁡2γ=

tan⁡(α+β−γ)tan⁡(α−β+γ)=tan⁡γtan⁡β⇒ sin⁡(α+β−γ)cos⁡(α−β+γ)sin⁡(α−β+γ)cos⁡(α+β−γ)=sin⁡γcos⁡βsin⁡βcos⁡γApplying componendo and dividendo, we get sin⁡2αsin⁡2(β−γ)=sin⁡(γ+β)sin⁡(γ−β)⇒sin⁡2(β−γ)sin⁡(β+γ)+sin⁡2αsin⁡(β−γ)=0⇒sin⁡(β−γ)(2cos⁡(β−γ)sin⁡(β+γ)+sin⁡2α)=0⇒sin⁡(β−γ)(sin⁡2α+sin⁡2β+sin⁡2γ)=0⇒sin⁡2α+sin⁡2β+sin⁡2γ=0⇒  (as β≠γ)