Let Δ = Expanding along C 3 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 sin 2 β – sinα cos 2 β) – 0 + cosα (cosα cos 2 β + cosα sin 2 β) Δ = sinα × sinα (sin 2 β + cos 2 β) + cosα × cosα (cos 2 β + sin 2 β) [Taking –sinα and cosα common] Since, sin 2 β + cos 2 β = 1 ∴ Δ = sin 2 α + cos 2 α Δ = 1 [sin 2 α + cos 2 α = 1]

Evaluate

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]