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Question

Prove that tan12A(ab)(ac)+tan12B(bc)(ba)+tan12C(ca)(cb)=1S.

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Solution

Putting tan12A= ((sb)(sc)s(sa)) etc.,
We have
=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ((sb)(sc)s(sa))(ab)(ac)+ ((sc)(sa)s(sb))(bc)(ba)+ ((sa)(sb)s(sc))(ca)(cb)⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥
=(cb)(sb)(sc)+(ac)(sc)(sa)+(ba)(sa)(sc)(ab)(bc)(ca)[s(sa)(sb)(sc)]
Nr={(bc){s2s(b+c)+bc}
=s2(bc)+s(b2c2)bc(bc)
=0+0(bc)(ca)(ab)
Note that on simplification,
bc(bc)+ca(ca)+ab(ab)
=(bc)(ca)(ab)]
Substituting in (1), we gat
L.H.S=NrDr=1[s(sa)(sb)(sc)]=1S

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