Thread: Please help me to do this problem!

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

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
Then
And:


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

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

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
Your result immediately follows
You can prove Napier Analogy by using Sine Law

Can you show a bit more detail? I havn't seen this method before...

Originally Posted by chancey

Can you show a bit more detail? I havn't seen this method before...

Have you tried?

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
Your result immediately follows
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

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

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.

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?

Originally Posted by malaygoel

COULD YOU PROVE THE SINE RULE?

In fact we can do better.
The extended sines law