Posted by **seema** on Tuesday, June 14, 2011 at 5:17am.

1. if sin 2A = Psin2B,

prove that tan(A+B)/tan(A-B)=P+1/P-1.

- Some Other Opinion ? - trig -
**Reiny**, Tuesday, June 14, 2011 at 8:42am
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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