If in a triangle ABC, sinA, sinB, sinC are in AP then show that

tan A/2 tan C/2 = 1/3

I have solved above problem as follows:

2b = a + c (Given)

2b + b = a + c + b

3b = 2s

3b + s = 2s + s

3(s - b) = s

(s - b)/s = 1/3

Since (s - b)/s = tan A/2 tan C/2

It is proved that tan A/2 tan C/2 = 1/3

My question is can we solve the problem starting from 2 sinB = sinA + sinC and without using 2b = a + c

Dear student,

No you have to use 2b=a+c anyhow to prove it...

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Sagar Singh

B.Tech, IIT Delhi