       Re: Inverse Tangent function

• To: mathgroup at smc.vnet.net
• Subject: [mg80023] Re: [mg79970] Inverse Tangent function
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Fri, 10 Aug 2007 06:46:19 -0400 (EDT)
• References: <200708100538.BAA10433@smc.vnet.net>

```On 10 Aug 2007, at 07:38, Boen S. Liong wrote:

>
> Let w1 = tan-1(x)  i.e. inverse tangent of x or tan(w1) = x
>
> w2 = tan-1(y)  i.e. tan(y) = w2
>
> and w3 = w2- w1
>
> problem: find tan(w3) in term of x and y.
>
> or better yet:
>
> let x= x1/z1
>
> and y= y1/z1
>
> problem:find tan(w3) in term of x and y.
>
> above. thank you.
>
> Boen S. Liong
>
>

TrigExpand[Tan[ArcTan[x] - ArcTan[y]]]
x/(Sqrt[x^2 + 1]*Sqrt[y^2 + 1]*((x*y)/(Sqrt[x^2 + 1]*Sqrt[y^2 + 1]) +
1/(Sqrt[x^2 + 1]*Sqrt[y^2 + 1]))) -
y/(Sqrt[x^2 + 1]*Sqrt[y^2 + 1]*((x*y)/(Sqrt[x^2 + 1]*Sqrt[y^2 +
1]) +
1/(Sqrt[x^2 + 1]*Sqrt[y^2 + 1])))

P.S. I don't see the point of the "or better yet" part of your question.

Andrzej Kozlowski

```

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