1
Question
If x=sin(α−β).sin(γ−δ);y=sin(β−γ).sin(α−δ) and z=sin(γ−α).sin(β−δ,) then
If x=sin(α−β).sin(γ−δ);y=sin(β−γ).sin(α−δ) and z=sin(γ−α).sin(β−δ,) then
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Solution
The correct option is A x+y+z=0
Given
x=sin(α−β)sin(γ−δ)
12(2sin(α−β)sin(γ−δ))
⟹12(cos(α−β−γ+δ))−cos(α−β+γ−δ)⋯(1)
y=sin(β−γ)sin(α−γ)
12(2sin(β−γ)sin(α−γ))
12(cos(β−γ−α+δ)−cos(α+β−γ−δ))
12(cos(α−β+γ−δ)−cos(α+β−γ−δ))⋯(2)
z=sin(γ−α)sin(β−γ)
⟹12(2sin(γ−α)sin(β−γ))
⟹12(cos(γ−α−β+δ)−cos(γ−α+β−δ))
⟹12(cos(α+β−γ−δ)−cos(α−β−γ+δ))⋯(3)
From (1)(2)(3)
⟹x+y+z
⟹12(cos(α−β−γ+δ))−cos(α−β+γ−δ)
+12(cos(α−β+γ−δ)−cos(α+β−γ−δ))12(cos(α+β−γ−δ)−cos(α−β−γ+δ))
⟹0
Given
x=sin(α−β)sin(γ−δ)
12(2sin(α−β)sin(γ−δ))
⟹12(cos(α−β−γ+δ))−cos(α−β+γ−δ)⋯(1)
y=sin(β−γ)sin(α−γ)
12(2sin(β−γ)sin(α−γ))
12(cos(β−γ−α+δ)−cos(α+β−γ−δ))
12(cos(α−β+γ−δ)−cos(α+β−γ−δ))⋯(2)
z=sin(γ−α)sin(β−γ)
⟹12(2sin(γ−α)sin(β−γ))
⟹12(cos(γ−α−β+δ)−cos(γ−α+β−δ))
⟹12(cos(α+β−γ−δ)−cos(α−β−γ+δ))⋯(3)
From (1)(2)(3)
⟹x+y+z
⟹12(cos(α−β−γ+δ))−cos(α−β+γ−δ)
+12(cos(α−β+γ−δ)−cos(α+β−γ−δ))12(cos(α+β−γ−δ)−cos(α−β−γ+δ))
⟹0
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