Given a triangle ABC let

X=tan[(B-C)/2]tan(A/2)

Y=tan[(C-A)/2]tan(B/2)

Z=tan[(A-B)/2]tan(C/2)

Prove that X+Y+Z+XYZ=0

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- May 10th 2006, 09:26 PM #1

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## Proof of NApier Analogy

Originally Posted by**malaygoel**

a/sinA =b/sinB =c/sinC =2R

(b-c)/(b+c)=(2RsinB-2RsinC)/(2RsinB+2RsinC)

=(sinB-sinC)/(sinB+sinC)

I can't write next step, use addition and subtraction fromulae and you will get X

- May 31st 2006, 03:55 AM #8

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- June 1st 2006, 09:37 AM #10

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Originally Posted by**malaygoel**

The extended sines law