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Question

Prove that ∣ ∣1+cosBcosC+cosBcosBcosC+cosA1+cosAcosA1+cosB1+cosA1∣ ∣

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Solution

We have,
D=∣ ∣1+cosBcosC+cosBcosBcosC+cosA1+cosAcosA1+cosB1+cosA1∣ ∣

D=∣ ∣cosB1cosC+cosBcosBcosC+cosAcosA1cosAcosB1cosA11∣ ∣

Applying R1R1+R2

D=∣ ∣(cosA+cosB+cosC1)(cosA+cosB+cosC1)cosA+cosBcosC+cosA1+cosAcosA1+cosB1+cosA1∣ ∣

Applying C1C1C2 and C2C2C1

D=∣ ∣00cosA+cosB1+cosC(1+cosC)cosAcosBcosAcosAcosB1∣ ∣

D=∣ ∣ ∣00cosA+cosB1+cosC(1+cosC)cosA(cosAcosB)cosAcosB1∣ ∣ ∣

On expanding w.r.t R1, we get
D=(cosA+cosB)[(1+cosC)(cosAcosB)(1+cosC)(cosAcosB)]

D=(cosA+cosB)×0
D=0

Hence, this is the answer.

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