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take the numerator of L.H.S
1-cosA+cosB-cos(A+B) simplifying 1-cosA = 2sin2 A/2 by half angle identity
cosB-cos(A+B) = 2sin(A/2)sin((A+2B)/2) difference into product identity
2sin2 A/2 + 2sin(A/2)sin((A+2B)/2) by simple trigonometric identities
by further simplifying
2sin(A/2)(sin A/2 + sin((A+2B)/2)
now applying sum to product identity
2sin(A/2)(2sin((A+B)/2)cosB/2-------------1
and then take denominator
1+cosA-cosB-cos(A+B)
similarly by half angle and sum into product identity we get
2cos2 A/2 + 2cos(A/2)cos((A+2B)/2)
2cos(A/2)(cos A/2 - cos((A+2B)/2)
applying difference to product identity we get
2cos(A/2)(2sin((A+B)/2)sinB/2---------------2
now dividing 1 by 2
we get
tanA/2cotB/2
Regards
Arun (askIITians forum expert)
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