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

```							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) ```
8 years ago
```							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
```
8 years ago
```							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)
```
one year ago
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