If in a triangle ABC, sinA, sinB, sinC are in AP then show thattan A/2 tan C/2 = 1/3I have solved above problem as follows:2b = a + c (Given)2b + b = a + c + b3b = 2s3b + s = 2s + s3(s - b) = s(s - b)/s = 1/3Since (s - b)/s = tan A/2 tan C/2It is proved that tan A/2 tan C/2 = 1/3My 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...

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..

Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi