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Question

If a2+b2+c2=1, then the value of
∣ ∣ ∣a2+(b2+c2)cosϕab(1cosϕ)ac(1cosϕ)ba(1cosϕ)b2+(c2+a2)cosϕbc(1cosϕ)ca(1cosϕ)cb(1cosϕ)c2+(a2+b2)cosϕ∣ ∣ ∣ is:

A
sinϕ
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B
sin2ϕ
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C
cosϕ
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D
cos2ϕ
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Solution

The correct option is D cos2ϕ
Multiplying C1 by a, C2 by b and C3 by c, we have
Δ=1abc∣ ∣ ∣a3+a(b2+c2)cosϕab2(1cosϕ)ac2(1cosϕ)ba2(1cosϕ)b3+b(c2+a2)cosϕbc2(1cosϕ)ca2(1cosϕ)cb2(1cosϕ)c3+c(a2+b2)cosϕ∣ ∣ ∣

Now take a,b and c common fromR1, R2 and R3 respectively, we have:

Δ=abcabc∣ ∣ ∣a2+(b2+c2)cosϕb2(1cosϕ)c2(1cosϕ)a2(1cosϕ)b2+(c2+a2)cosϕc2(1cosϕ)a2(1cosϕ)b2(1cosϕ)c2+(a2+b2)cosϕ∣ ∣ ∣

Applying C1C1+C2+C3, we get

Δ=abcabc∣ ∣ ∣1b2(1cosϕ)c2(1cosϕ)1b2+(c2+a2)cosϕc2(1cosϕ)1b2(1cosϕ)c2+(a2+b2)cosϕ∣ ∣ ∣

[a2+b2+c2=1]
Applying R2R2R1,R3R3R1

Δ=∣ ∣ ∣1b2(1cosϕ)c2(1cosϕ)0(a2+b2+c2)cosϕ000(a2+b2+c2)cosϕ∣ ∣ ∣

Expanding along C1 we get Δ=cosϕ00cosϕ=cos2ϕ

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