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Question

If A,B,C are the angles of a triangle, then the value of determinant


∣∣ ∣∣−1+cosBcosC+cosBcosBcosC+cosA−1+cosAcosA−1+cosB−1+cosA−1∣∣ ∣∣ is

A
0
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B
1
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C
1
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D
2
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Solution

The correct option is B 0
Given ∣ ∣1+cosBcosC+cosBcosBcosC+cosA1+cosAcosA1+cosB1+cosA1∣ ∣

=∣ ∣1cosCcosBcosC1cosAcosBcosA1∣ ∣(C1C1C3C2C2C3)

=1a∣ ∣acosCcosBacosC1cosAacosBcosA1∣ ∣

=1a∣ ∣0cosCcosB01cosA0cosA1∣ ∣(C1C1+bC2+cC3)

=0

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