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Question

If ϕ(α,β)=∣ ∣cosαsinα1sinαcosα1cos(α+β)sin(α+β)1∣ ∣, then

A
ϕ(300,200)=ϕ(400,200)
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B
ϕ(200,400)=ϕ(200,600)
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C
ϕ(100,200)=ϕ(200,200)
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D
None of the above
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Solution

The correct options are
A ϕ(300,200)=ϕ(400,200)

C ϕ(100,200)=ϕ(200,200)

ϕ(α,β)=∣ ∣cosαsinα1sinαcosα1cos(α+β)sin(α+β)1∣ ∣

Applying R3R3R1(cosβ)+R2(sinβ)
=∣ ∣cosαsinα1sinαcosα1001+sinβcosβ∣ ∣

=(1+sinβcosβ)(cos2α+sin2α)
=1+sinβcosβ which is independent of α.

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