Inverse tangent

Let w1 = tan-1(x) i.e. tan(x) = w1
w2 = tan-1(y) i.e. tan(y) = w2
w3 = w1 - w2
problem what is tan(w3) in terms of x and y.
or:
Let w1 = tan-1(x1/z1) i.e. tan(w1) = x1/z1
w2 = tan-1(x2/z2) i.e. tan(w2) = x2/z2
w3 = w1 -w2
what is tan(w3) in term of of x1, x2, z1, z2?
Thank you,
Boen S. Liong
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On Thu, 9 Aug 2007 12:19:12 +0100, Boen S. Liong wrote

Let Z be the base length of both triangles, then :
tan(w3) = (x1 - x2).cos(w1) / { Sqrt(z^2 + y2^2).cos(w1-w2) }
AAR
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On Thu, 9 Aug 2007 12:19:12 +0100, Boen S. Liong wrote

Sorry, previous post should read :
Let Z be the base length of both triangles, then :
tan(w3) = (x1 - x2).cos(w1) / { Sqrt(z^2 + x2^2).cos(w1-w2) }
AAR
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On Sun, 12 Aug 2007 18:03:39 +0100, AAR wrote

Or, in terms of x1, x2 and Z only :
tan(w3) = Z(x1-x2) / (Z^2 + x1x2)
AAR
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AAR wrote:
...

Can you tie that back to arctangent? That was the original question.
Jerry
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On Mon, 13 Aug 2007 15:32:39 +0100, Jerry Avins wrote

The original question was "What is tan(w3) in terms of x1, x2, z1, z2 ?" So it seems to me that I've given what was asked for.
Note: there's no need for two z's because the triangles have a common base length which I call Z.
AAR
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AAR wrote:

The subject of this thread: Inverse tangent. The original query begins:
Let w1 = tan-1(x) i.e. tan(x) = w1
w2 = tan-1(y) i.e. tan(y) = w2
I guess I was misled.
Jerry
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