54 54 o o o Share with your friends Share 0 Neha Sethi answered this Dear student sinαcosαsinα+δsinβcosβsinβ+δsinγcosγsinγ+δ=sinαcosαsinαcosδ+cosαsinδsinβcosβsinβcosδ+cosβsinδsinγcosγsinγcosδ+cosγsinδ=sinαcosαsinαcosδsinβcosβsinβcosδsinγcosγsinγcosδ+sinαcosαcosαsinδsinβcosβcosβsinδsinγcosγcosγsinδTaking cosδ common from C3 in first determinant and sinδ common from C3 in second determinant=cosδsinαcosαsinαsinβcosβsinβsinγcosγsinγ+sinδsinαcosαcosαsinβcosβcosβsinγcosγcosγAs C1 and C3 are same in first detreminant and C2 and C3 are samein second determinant,hence they both will be zero.=cosδ×0+sinδ×0=0+0=0⇒sinαcosαsinα+δsinβcosβsinβ+δsinγcosγsinγ+δ=0 Regards -1 View Full Answer Prakash answered this Please find this answer 0
