# trig

1. if sin 2A = Psin2B,
prove that tan(A+B)/tan(A-B)=P+1/P-1.

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1. I have assumed the RS is (P+1)/(P-1) 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)/(P-1)
= [(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(A-B) 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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