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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


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


Grade:12

1 Answers

SAGAR SINGH - IIT DELHI
879 Points
10 years ago

Dear student,

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

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Sagar Singh

B.Tech, IIT Delhi

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