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Question

The determinant ∣ ∣sinαcosα1sinβcosβ1sinγcosγ1∣ ∣ is equal to

A
4sinαβ2sinαγ2sinγα2
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B
sinα+sinβ+sinγ
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C
sin(αβ)+sin(βγ)+sin(γα)
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D
none of these
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Solution

The correct options are
A 4sinαβ2sinαγ2sinγα2
C sin(αβ)+sin(βγ)+sin(γα)
Expanding along with C3
∣ ∣sinαcosα1sinβcosβ1sinγcosγ1∣ ∣

=1(sinβcosγsinγcosβ)1(sinαcosγsinγcosα)+1(sinαcosβsinβcosα)

=sin(βγ)+sin(γα)+sin(αβ)
Now let A=(βγ),B=(γα),C=(αβ)
we have A+B+C=0
Now
sin(βγ)+sin(γα)+sin(αβ)=sinA+sinB+sinC

=2sinA+B2cosAB2+2sinC2cosC2

=2sinC2cosAB2+2sinC2cos(A+B)2

=2sinC2(cosAB2cosA+B2)

=2sinC2(2sinA2sinB2)

=4sinA2sinB2sinC2

4sin(βγ)2sin(γα)2sin(αβ)2

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