In triangle ABC, (a + b + c) (tan A/2+tanB/2)=
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Answer:
→tanA/2,tanB/2,tanC/2 are in HP
⇒
tanA/2
1
,
tanB/2
1
,
tanC/2
1
are in A.P
⇒
tanB/2
2
=
tanA/2
1
+
tanC/2
1
⇒2cotB/2=cotA/2+cotC/2
⇒2cot(
2
π−(A+C)
)=cotA/2+cotC/2
⇒2tan(
2
A+C
)=cotA/2+cotC/2
⇒2[
1−tanA/2tanC/2
tanA/2+tanC/2
]=
tanA/2tanC/2
tanA/2+tanC/2
let tanA/2tanC/2=y
⇒
1−y
2
=
y
1
⇒2y=1−y
⇒y=−1/3
⇒
cotA/2cotC/2=
y
1
=3
⇒(C)
Step-by-step explanation:
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