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Question

# The value of the determinant is ∣∣ ∣ ∣∣cos(A−P)cos(A−Q)cos(A−R)cos(B−P)cos(B−Q)cos(B−R)cos(C−P)cos(C−Q)cos(C−R)∣∣ ∣ ∣∣

A
zero for all values of P, Q, R, A, B, and C.
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B
one for all values of P, Q, R, A, B, and C.
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C
zero only if P= Q= R= A= B= C.
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D
one only if P= Q= R= A= B= C.
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Solution

## The correct option is A zero for all values of P, Q, R, A, B, and C.Let Δ=∣∣ ∣ ∣∣cos(A−P)cos(A−Q)cos(A−R)cos(B−P)cos(B−Q)cos(B−R)cos(C−P)cos(C−Q)cos(C−R)∣∣ ∣ ∣∣ ⇒ Δ=∣∣ ∣ ∣∣cos A cos P+sin A sin Pcos(A−Q)cos(A−R)cos B cos P+sin B sin Pcos(B−Q)cos(B−R)cos C cos P+sin C sin Pcos(C−Q)cos(C−R)∣∣ ∣ ∣∣ ⇒ Δ=∣∣ ∣ ∣∣cos A cos Pcos(A−Q)cos(A−R)cos B cos Pcos(B−Q)cos(B−R)cos C cos Pcos(C−Q)cos(C−R)∣∣ ∣ ∣∣+∣∣ ∣ ∣∣sin A sin Pcos(A−Q)cos(A−R)sin B sin Pcos(B−Q)cos(B−R)sin C sin Pcos(C−Q)cos(C−R)∣∣ ∣ ∣∣ ⇒ Δ=cos P∣∣ ∣ ∣∣cos Acos(A−Q)cos(A−R)cos Bcos(B−Q)cos(B−R)cos Ccos(C−Q)cos(C−R)∣∣ ∣ ∣∣+sin P∣∣ ∣ ∣∣sin Acos(A−Q)cos(A−R)sin Bcos(B−Q)cos(B−R)sin Ccos(C−Q)cos(C−R)∣∣ ∣ ∣∣ Applying C2→C2−C1 cos Q, C3→C3−C1 cos R in first determinant and C2→C2−C1 sin Q and in second determinant ⇒ Δ=cos P∣∣ ∣∣cos Asin A sin Qsin A sin Rcos Bsin B sin Qsin B sin Rcos Csin C sin Qsin C sin R∣∣ ∣∣+sin P∣∣ ∣∣sin Acos A cos Qcos A cos Rsin Bcos B cos Qcos B cos Rsin Ccos C cos Qcos C cos R∣∣ ∣∣Δ=cos P sin Q sin R∣∣ ∣∣cos Asin Asin Acos Bsin Bsin Bcos Csin Csin C∣∣ ∣∣+sin P cos Q cos R∣∣ ∣∣sin Acos Acos Asin Bcos Bcos Bsin Ccos Ccos C∣∣ ∣∣ Δ = 0 + 0 = 0

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