If A,B,C,P,Q,R∈R and 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)

WriteΔ=cos⁡Asin⁡A0cos⁡Bsin⁡B0cos⁡Csin⁡C0cos⁡P−sin⁡P0cos⁡Q−sin⁡Q0cos⁡R−sin⁡R0=0Alternate SolutionWrite Δ=cos⁡PΔ1−sin⁡PΔ2write Δ1=cos⁡A cos⁡(A+Q) cos⁡(A+R)cos⁡B cos⁡(B+Q) cos⁡(B+R)cos⁡C cos⁡(C+Q) cos⁡(C+R)and Δ2=sin⁡A cos⁡(A+Q) cos⁡(A+R)sin⁡B cos⁡(B+Q) cos⁡(B+R)sin⁡C cos⁡(C+Q) cos⁡(C+R)In Δ1,use C2→C2−cos⁡QC1C3→C3−cos⁡RC1Δ1=cos⁡A sin⁡Asin⁡Q sin⁡Asin⁡Rcos⁡B sin⁡Bsin⁡Q sin⁡Bsin⁡Rcos⁡C sin⁡Csin⁡Q sin⁡Csin⁡R=0 ∵C2 and C3 are propotional Similarly, Δ2=0Thus ∆ = 0

If A,B,C,P,Q,R∈R and 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)

WriteΔ=cos⁡Asin⁡A0cos⁡Bsin⁡B0cos⁡Csin⁡C0cos⁡P−sin⁡P0cos⁡Q−sin⁡Q0cos⁡R−sin⁡R0=0Alternate SolutionWrite Δ=cos⁡PΔ1−sin⁡PΔ2write Δ1=cos⁡A cos⁡(A+Q) cos⁡(A+R)cos⁡B cos⁡(B+Q) cos⁡(B+R)cos⁡C cos⁡(C+Q) cos⁡(C+R)and Δ2=sin⁡A cos⁡(A+Q) cos⁡(A+R)sin⁡B cos⁡(B+Q) cos⁡(B+R)sin⁡C cos⁡(C+Q) cos⁡(C+R)In Δ1,use C2→C2−cos⁡QC1C3→C3−cos⁡RC1Δ1=cos⁡A sin⁡Asin⁡Q sin⁡Asin⁡Rcos⁡B sin⁡Bsin⁡Q sin⁡Bsin⁡Rcos⁡C sin⁡Csin⁡Q sin⁡Csin⁡R=0 ∵C2 and C3 are propotional Similarly, Δ2=0Thus ∆ = 0

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If $A, B, C, P, Q, R \in R$ and $Δ = |\begin{array}{lll} \cos (A + P) \cos (A + Q) \cos (A + R) \\ \cos (B + P) \cos (B + Q) \cos (B + R) \\ \cos (C + P) \cos (C + Q) \cos (C + R) \end{array}|$

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$Δ$ depends on $A, B, C$

$Δ$ depends on $A, B, C, P, Q, R$

$Δ$ depends on $P, Q, R$

none of these

answer is D.

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Detailed Solution

Write

$\begin{array}{l} Δ = |\begin{array}{lll} \cos A & \sin A & 0 \\ \cos B & \sin B & 0 \\ \cos C & \sin C & 0 \end{array}| |\begin{array}{ccc} \cos P & - \sin P & 0 \\ \cos Q & - \sin Q & 0 \\ \cos R & - \sin R & 0 \end{array}| \\ = 0 \end{array}$

Alternate Solution

Write $Δ = \cos P Δ_{1} - \sin P Δ_{2}$

write $Δ_{1} = |\begin{array}{lll} \cos A \cos (A + Q) \cos (A + R) \\ \cos B \cos (B + Q) \cos (B + R) \\ \cos C \cos (C + Q) \cos (C + R) \end{array}|$

and $Δ_{2} = |\begin{array}{lll} \sin A \cos (A + Q) \cos (A + R) \\ \sin B \cos (B + Q) \cos (B + R) \\ \sin C \cos (C + Q) \cos (C + R) \end{array}|$

In $Δ_{1},$ use $C_{2} \to C_{2} - \cos Q C_{1}$

$C_{3} \to C_{3} - \cos R C_{1}$

$Δ_{1} = |\begin{array}{lll} \cos A \sin A \sin Q \sin A \sin R \\ \cos B \sin B \sin Q \sin B \sin R \\ \cos C \sin C \sin Q \sin C \sin R \end{array}|$