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Question

Evaluate : Δ=∣ ∣0sinαcosαsinα0sinβcosαsinβ0∣ ∣

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Solution

We have, Δ=∣ ∣0sinαcosαsinα0sinβcosαsinβ0∣ ∣

On expanding along first row, we get

Δ=00sinβsinβ0sinαsinαsinβcosα0cosαsinα0cosαsinβ

Δ=sinα(0sinβcosα)cosα(sinαsinβ)

Δ=sinαcosαsinβsinαcosαsinβ

Δ=0

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