Posted by seema on Tuesday, June 14, 2011 at 5:17am.
1. if sin 2A = Psin2B,
prove that tan(A+B)/tan(AB)=P+1/P1.

Some Other Opinion ?  trig  Reiny, Tuesday, June 14, 2011 at 8:42am
I have assumed the RS is (P+1)/(P1) instead of the way it was typed.
I have been working on this off and on since it was first posted yesterday and seem to be getting nowhere.
from sin 2A = Psin 2B
2sinAcosA = 2PsinBcosB
P = sinAcosA/(sinBcosB)
I then went to the right side of the identity
RS = (P+1)/(P1)
= [(sinAcosA + sinBcosB)/(sinBcosB)] / [(sinAcosA  sinBcosB)/(sinBcosB)
= (sinAcosA + sinBcosB)/(sinAcosA  sinBcosB)
Now, if this truly is an identity, it should be equal to
tan(A+B)/tan(AB) for all values of A and B, which would make the tangent defined.
So I let A = 40° and B = 10°
LS = tan50/tan30 = appr. 2.06
RS = (sin50cos50 + sin10cos10)/(sin50cos50  sin10cos10) = appr. 0.321
LS ≠ RS, so it is not an identity.
Other thoughts are appreciated.
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