Sides a , b , c of `!ABC` are in A.P. and `costheta_(1)=a/(b+c)costheta_(2)=b/(a+c),costheta_(3)=c/(a+b)` , then `tan^(2)(theta_(1))/2+tan^(2)(theta_(3))/2`=

Options

A. `2//3`
B. 1
C. `sqrt5//3`
D. none of these

Explanation

We have,
`costheta_(1)=a/(b+c)` and `costheta_(3)=c/(a+b)`
`thereforetan^(2)(theta_(1))/2=(1-costheta_(1))/(1+costheta_(1))` and `tan^(2)(theta_(3))/2=(1-costheta_(3))/(1+costheta_(3))`
`rArrtan^(2)(theta_(1))/2=(b+c-a)/(a+b+c)` and `tan^(2)theta_(3)=(a+b-c)/(a+b+c)`
`rArrtan^(2)(theta_(1))/2+tan^(2)(theta_(3))/2=(2b)/(a+b+c)`
`rArrtan^(2)(theta_(1))/2+tan^(2)(theta_(3))/2=(2b)/(3b)=2/3` `[because` a,b,c are in A.P.
`thereforea+c=2b]`