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

2. Originally Posted by mpgc_ac
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
Some formulae that might help:
If $t = \tan \frac{\theta}{2}$
Then $\tan \theta = \frac{2 t}{1 - t^2}$
And:
$tan(\theta - \phi) = \frac{\tan \theta - \tan \phi}{1+\tan \theta \tan \phi}$

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

3. 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
You can prove Napier Analogy by using Sine Law

4. Originally Posted by malaygoel
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
You can prove Napier Analogy by using Sine Law
Can you show a bit more detail? I havn't seen this method before...

5. Originally Posted by chancey
Can you show a bit more detail? I havn't seen this method before...
Have you tried?

6. Originally Posted by malaygoel
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
You can prove Napier Analogy by using Sine Law
X=(b-c)/(b+c)
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

7. ## Proof of NApier Analogy

Originally Posted by malaygoel
X=TAN((B-C)/2)TAN(A/2)=(b-c)/(b+c) [NAPIER ANALOGY:a,b,c are sides of triangle]
You can prove Napier Analogy by using Sine Law
sine law states that
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

8. Originally Posted by malaygoel
sine law states that
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
Ahhh... I sort of get it, and I learnt a new sine rule, cool thanks.

9. Originally Posted by chancey
Ahhh... I sort of get it, and I learnt a new sine rule, cool thanks.
COULD YOU PROVE THE SINE RULE?

10. Originally Posted by malaygoel
COULD YOU PROVE THE SINE RULE?
In fact we can do better.
The extended sines law