:confused: 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 PMmpgc_acPlease help me to do this problem!
:confused: 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 - May 27th 2006, 03:15 AMchanceyQuote:

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 AMmalaygoel
X=TAN((B-C)/2)TAN(A/2)=(b-c)/(b+c) [NAPIER ANALOGY:a,b,c are sides of triangle]

Similar are Y and Z

Then solve (X+Y)/XY+1

You will find it is equal to –Z

Your result immediately follows

You can prove Napier Analogy by using Sine Law - May 30th 2006, 04:46 AMchanceyQuote:

Originally Posted by**malaygoel**

- May 30th 2006, 07:07 AMmalaygoelQuote:

Originally Posted by**chancey**

- May 30th 2006, 07:14 AMmalaygoelQuote:

Originally Posted by**malaygoel**

Y=(c-a)/(c+a)

(X+Y)/(XY+1)={(b-c)/(b+c)+(c-a)/(c+a)}/[{(c-a)/(c+a)}*{(b-c)/(b+c)}+1]

=(b-a)/(b+a)=-Z

therefore,

(X+Y)/(XY+1)=-Z

multiply both sides by XY+1

X+Y=-Z-XYZ

X+Y+Z+XYZ=0

Good Luck :) - May 30th 2006, 07:20 AMmalaygoelProof of NApier AnalogyQuote:

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 AMchanceyQuote:

Originally Posted by**malaygoel**

- Jun 1st 2006, 05:21 AMmalaygoelQuote:

Originally Posted by**chancey**

- Jun 1st 2006, 08:37 AMThePerfectHackerQuote:

Originally Posted by**malaygoel**

The extended sines law