if tanA/2,tanB/2,tanC/2 are in harmonic progression then find value of cotB/2

Dear student,

Use the concept:

tan (A + B + C) =

=

=

So, since tan (A + B + C) = tan (180⁰) = 0

So, tanA + tanB + tanC - tanA + tanB + tanC = 0

=> tanA + tanB + tanC = tanA + tanB + tanC.
(vi) cotA + cotB + cotA + cotC + cotB cotC = 1
(vii)
(viii)

tanA/2 , tanB/2 , tanC/2 are in HP so

 

2/tanB/2 = 1/tanA/2 + 1/tanC/2

2/tanB/2 = (tanA/2 + tanB/2)/tanA/2tanB/2            .....................1

 

cotB/2 =cot[pi/2-(A+C)/2]                                      { B/2 = pi/2 - (A+C/2) }

         = tan(A/2+C/2) = tan(A/2)+tan(C/2)/1-tan(A/2)tan(C/2)

 

from this eq value of tanA/2 + tanB/2 = cotB/2 [ 1-tanA/2tanC/2 ]        ....................2

plugging this in eq 1 we get

 

2/tanB/2 = cotB/2 [ 1-tanA/2tanC/2 ]/tanA/2tanC/2

tanA/2tanC/2 = 1/3

now using tangent formulas

this expression simplifies to

a+c = 2b

a,b,c are in Ap

now , cot B/2 can be calulated easily by using

cotB/2 = { s(s-b)/(s-a)(s-c) }^1/2

 

approve if u like my ans

Let TanA/2 =a TanB/2=bTanC/2=C2/b=1/a+1/C ............(1)A + B +C = 180A/2 +B/2 +C/2 =90ORTan(A/2+C/2)=cotB/2=1/b.....(2)From solving (1) & (2) we get TanA/2 ×TanC/2 =1÷3 From 1 we have a+c=2/3b......(3)Applying A.M.>=G.Ma+c /2>=root(ac)Solving we get 1/b or CotB/2>=root(3)Hence min. Cot B/2= root(3)