IN A TRIANGLE ABC , tan A/2=5/6 and tan c/2=2/5 then a)a.b.c are in A P b)cos A,cos b, cos C are in AP c)sin a,sin b,sin c are in GP d)none of these

tanA/2 = [(s-b)(s-c)/s(s-a)]^1/2                 ....................1

tanC/2 = [(s-a)(s-b)/s(s-c)]^1/2                ..................2

eq1 * eq2 = tanA/2tanB/2 = [s-b/s]              .................3

tanA/2 = 5/6          (given)

tanC/2 = 2/5          (given)

tanA/2tanB/2 = 5/6*2/5 = 1/3

from eq 3 , (s-b)/s = 1/3

    3s - 3b = s

    2s = 3b

   2(a+b+c)/2 = 3b

    a+c = 2b

a,b,c are in AP

approve if u like my ans

dear GAVEESH

a,b,c are in AP .........