In any Δ ABC, prove that
(i) (r 3 + r 1 ) (r 3 + r 2 ) sin C = 2 r 3 
(ii) 
(iii) (r + r 1 ) tan 
+ (r + r 2 ) tan 
+ (r + r 3 ) tan 
= 0
(iv) 
= 16R 2 – a 2 –b 2 – c 2 .
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) L.H.S. = (r 3 + r 1 ) (r 3 + r 2 ) sin C
= 
sin C
= 
sin C
= 
sin C
= 
= 
= 2 sr 3 
R.H.S. = 2r 3
= 2r 3 
= 2sr 3
∴ L.H.S. = R.H.S.
(ii) L.H.S. = – 
= – 
= 
= R.H.S.
(iii) First term = (r + r 1 ) tan 
= 
cot 
= 
. 
. 
= b – c
similarly second term= c – a
& third term = a – b
∴ L.H.S. = b – c + c – a + a – b = 0 = R.H.S.
(iv) r 1 + r 2 + r 3 – r = 4R
∴ (r 1 + r 2 + r 3 – r) 2 = r 12 + r 22 + r 32 + r 2 – 2r (r 1 + r 2 + r 3 ) + 2(r 1 r 2 + r 2 r 3 + r 3 r 1 ) ........(i)
r(r 1 + r 2 + r 3 ) = ab + bc + ca – s 2
and r 1 r 2 + r 2 r 3 + r 3 r 1 = s 2
∴ from equation (i)
16R 2 = r 2 + r 12 + r 22 + r 32 – 2 (ab + bc + ca – s 2 ) + 2s 2
∴ r 2 + r 12 + r 22 + r 32 = 16 R 2 – 4 s 2 + 2 (ab + bc + ca)
= 16R 2 – (a + b + c) 2 + 2 (ab + bc + ca)
= 16R 2 – a 2 – b 2 – c 2
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