Answer :

Let Δ =

Expanding along C3 we get


Δ = -sinα (-sinβ × sinα sinβ – cosβ × sinα cosβ) – 0 (sinα cosβ × cosα sinβ – cosα cosβ × sinα sinβ) + cosα (cosα cosβ × cosβ – cosα sinβ × (-sinβ))


Δ = -sinα (-sinα sin2β – sinα cos2β) – 0 + cosα (cosα cos2β + cosα sin2β)


Δ = sinα × sinα (sin2β + cos2β) + cosα × cosα (cos2β + sin2β)


[Taking –sinα and cosα common]


Since, sin2β + cos2β = 1


Δ = sin2α + cos2α


Δ = 1 [sin2α + cos2α = 1]


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