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

Answer 1

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.