In a Δ ABC, prove that
(i) 2 
= c + a – b.
(ii) 
+ 
+ 
= 
(iii) 4 
= (a + b + c) 2
(iv) (b – c) cot 
+ (c – a) cot 
+ (a – b) cot 
= 0
(v) 4 Δ (cot A + cot B + cot C) = a 2 + b 2 + c 2
(vi) 
. cos 
.cos 
. cos 
= Δ
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) L.H.S. = 2a sin 2 
+ 2 c sin 2 
= a(1 – cos c) + c(1 – cos A)
= a + c – (a cos C + c cos A)
= a + c – b
= R.H.S.
(ii) L.H.S. = 
+ 
+ 
= 
. 
+ 
. 
+ 
. 
= 
= 
.
(iii) L.H.S. = 2bc(1 + cos A) + 2ca(1 + cos B) + 2ab(1 + cos C)
= 2bc + 2ca + 2ab + 2bc cos A + 2ca cos B + 2 ab cos C
= 2 
+ a 2 + b 2 + c 2 = (a + b + c) 2 = R.H.S.
(iv) L.H.S. = (b – c) 
+ (c – a) 
+ (a – b) 
(b – c) cot 
= k(sin B – sin C) 
= 2k cos 
sin 
= 2k sin 
sin 
= k [cos C – cos B]
similarly (c – a) cot 
= k[cos A – cos C]
and (a – b) cot 
= k[cos B – cos A]
∴ L.H.S. = k[cos C – cos B + cos A – cos C + cos B – cos A]
= 0
= R.H.S.
(v) L.H.S. {k = 4 Δ (cot A + cot B + cot C)
= 4 Δ 
= 2bc cos A + 2 ca cos B + 2ab cos C
= a 2 + b 2 + c 2 = R.H.S. nk;ka i{k
(vi) L.H.S. ck;ka i{k = 
cos 
.cos 
.cos 
= 
= 
= Δ = R.H.S
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