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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
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