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Question
If A + B + C = 0, then prove that `|(1, cos"c", cos"B"),(cos"C", 1, cos"A"),(cos"B", cos"A", 1)|` = 0
Solution
L.H.S. = `|(1, cos"c", cos"B"),(cos"C", 1, cos"A"),(cos"B", cos"A", 1)|`
Expanding along C1
= `1|(1, cos"A"),(cos"A", 1)| - cos"C"|(cos"C", cos"B"),(cos"A", 1)| + cos"B"|(cos"C", cos"B"),(1, cos"A")|`
= 1(1 – cos2A) – cos C(cos C – cos A cos B) + cos B(cos A cos C – cos B)
= sin2A – cos2C + cos A cos B cos C + cos A cos B cos C – cos2B
= sin2A – cos2B – cos2C + 2 cos A cos B cos C
= – cos(A + B) · cos(A – B) – cos2C + 2 cos A cos B cos C .....[∵ sin2A – cos2B = – cos(A + B) · cos(A – B)]
= – cos(– C) · cos(A – B) + cos C(2 cos A cos B – cos C) .....[∵ A + B + C = 0]
= – cos C(cos A cos B + sin A sin B) + cos C(2 cos A cos B – cos C)
= – cos C(cos A cos B + sin A sin B – 2 cos A cos B + cos C)
= – cos C(– cos A cos B + sin A sin B + cos C)
= cos C(cos A cos B – sin A sin B – cos C)
= cos C[cos(A + B) – cos C]
= cos C[cos (– C) – cos C] .....[∵ A + B = – C]
= cos C[cos C – cos C]
= cos C · 0
= 0 R.H.S.
L.H.S. = R.H.S.
Hence proved.
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