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Question:

If Δ1=∣∣∣xsinθcosθ−sinθ−x1cosθ1x∣∣∣ and Δ2=∣∣∣xsin2θcos2θ−sin2θ−x1cos2θ1x∣∣∣, x≠0; then for all θ∈(0,π/2): Δ1−Δ2=x(cos2θ−cos4θ) Δ1+Δ2=−2;x3 Δ1−Δ2=−2;x3 Δ1+Δ2=−2;(x3+x−1;)

Δ1−Δ2=x(cos2θ−cos4θ)

Δ1−Δ2=−2;x3

Δ1+Δ2=−2;x3

Δ1+Δ2=−2;(x3+x−1;)

Solution:

Δ1=f(θ)=∣∣∣xsinθcosθ−sinθ−x1cosθ1x∣∣∣=−x3andΔ2=f(2θ)=∣∣∣xsin2θcos2θ−sin2θ−x1cos2θ1x∣∣∣=−x3SoΔ1+Δ2=−2;x3