Question

In AABC, Wb c = 3a then Tan Ba Tan 612 =​

Answer 1

Answer:

ANSWER

cot

2

B

⋅cot

2

C

we know,

cot

2

B

=

(s−a)(s−c)

s(s−b)

,cot

2

C

=

(s−a)(s−b)

s(s−c)

∴cot

2

B

⋅cot

2

C

=

s−a

s

s=

2

a+b+c

⇒

2

a+3a

,s=2a

cot

2

B

⋅cot

2

C

=

2a−a

2a

⇒2