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, 08:26 PM #1

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- May 27th 2006, 03:15 AM #2

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Originally Posted by**mpgc_ac**

If $\displaystyle t = \tan \frac{\theta}{2}$

Then $\displaystyle \tan \theta = \frac{2 t}{1 - t^2}$

And:

$\displaystyle tan(\theta - \phi) = \frac{\tan \theta - \tan \phi}{1+\tan \theta \tan \phi}$

I tried, but the algebra is too much, there must be another way

- May 30th 2006, 12:58 AM #3

- May 30th 2006, 04:46 AM #4

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- May 30th 2006, 07:07 AM #5

- May 30th 2006, 07:14 AM #6

- May 30th 2006, 07:20 AM #7
## 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, 02:55 AM #8

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- Jun 1st 2006, 05:21 AM #9

- Jun 1st 2006, 08:37 AM #10

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Originally Posted by**malaygoel**

The extended sines law