In a Δ ABC, prove that :
(i)a sin (B – C) + b sin (C – A) + c sin (A – B) = 0
(ii) 
+ 
+ 
= 0
(iii)2(bc cos A + ca cos B + ab cos C) = a 2 + b 2 + c 2
(iv)(a – b) 2 cos 2 
+ (a + b) 2 sin 2 
= c 2
(v)b 2 sin 2C + c 2 sin 2B = 2bc sin A
(vi) 
= 
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) L.H.S. = a sin (B – C) + b sin (C – A) + c sin (A – B)
= k sin A sin (B – C) + k sin B sin (C – A) + k sin C sin (A – B)
= k (sin 2 B – sin 2 C) + k (sin 2 C – sin 2 A) + k (sin 2 A – sin 2 B)
= 0 = R.H.S.
(ii) L.H.S. = 
first term = 
= 
= k 2 sin (B + C) sin (B – C)
= k 2 (sin 2 B – sin 2 C)
Similarly 
= k 2 (sin 2 C – sin 2 A)
and 
= k 2 (sin 2 A – sin 2 B)
∴ L.H.S. = k 2 (sin 2 B – sin 2 C + sin 2 C – sin 2 A + sin 2 A – sin 2 B)
= 0 = R.H.S.
(iii) L.H.S. = 2bc cos A + 2ca cos B + 2ab cos C
= b 2 + c 2 – a 2 + a 2 + c 2 – b 2 + a 2 + b 2 – c 2
= a 2 + b 2 + c 2
= R.H.S
(iv) L.H.S. = a 2 
– 2ab 
= a 2 + b 2 – 2ab cos C
= a 2 + b 2 – (a 2 + b 2 – c 2 )
= c 2 = R.H.S.
(v) L.H.S. = b 2 sin 2C + c 2 sin 2B
= 2b 2 sin C cos C + 2c 2 sin B cos B
= 2k 2 sin 2 B cos C sin C + 2k 2 sin 2 C sin B cos B ( b = ksin B, c = ksin C)
= 2k 2 sin B sin C [sin B cos C + cos B sin C]
= 2(k sin B) (k sin C) sin (B + C)
= 2bc sin A
(vi) R.H.S = 
c = a cos B + b cos A,
b = c cos A + a cos C
= 
= 
= 
= L.H.S.
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