Question

in triangle abc ,tan (a+b/2) tan c/2​

Answer 1

Answer:

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)