If in ABC triangle 3(tanA/2 +tanC/2)=2cotB/2 then prove that the sides of triangle a,b,c are in A.P
Answers
Answer:
tanA/2,tanB/2,tanC/2 in HP
⇒cotA/2,cotB/2,cotC/2 in AP
⇒2cotB/2=cotA/2+cotC/2
⇒
△
2S(S−b)
=
△
S(S−a)
+
△
S(S−c)
⇒2(S−b)=S−a+S−c
⇒2b=a+b
cosB=
2ac
c
2
+a
2
−b
2
=
2ac
c
2
+a
2
−(
2
a+c
)
2
cosB=
8ac
4c
2
+4a
2
−(a
2
+c
2
+2ac)
1+tan
2
(B/2)
1−tan
2
B/2
=
8ac
3c
2
+3a
2
−2ac
cot
2
B/2+1
1−cot
2
B/2−1
=
8a
3c
+
8c
3a
−
8
2
cot
2
B/2+1
cot
2
B/2−1
−1−
8a
3c
+
8c
3a
−
8
2
−1
⇒
cot
2
(B/2)+1
−2
=3(
8a
c
+
8c
a
−
12
5
)
1+cot
2
(B/2)=
3(
8c
a
+
8a
c
−
12
5
)
−2
cot
2
B/2=
3(
8c
a
+
8a
c
−
12
5
)
−2
−1
(cot
2
B/2)=
3(
8
1
(
c
a
+
a
c
)−
12
5
)
−2
−1 [∵x+
x
1
≥2]
(cot
2
B/2)
min
=
3(
8
1
.2−
12
5
)
−2
−1
=
3(
12
−2
)
−2
−1
=4−1=3
⇒(cotB/2)
min
=
3
Answer By
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Toppr
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