# Trig to algrebraic translation of a particular equation

I solved a partial derivative problem and have the correct answer but the book I am using, Courant's Differential and Integral Calculus, has the answer in algebraic rather than trigonometric form. I never really worked on learning the translation of trig equations to algebraic form (except in the most basic cases). Anyway, my answer is in the form:

$$\frac{\sec^2(\arctan(x)+\arctan(y))}{(x^2+1)}$$

Courant has:

$$\frac{1+y^2}{(1-xy)^2}$$

I'd appreciate if someone could show me the translation presumably using right triangles and also perhaps suggest a good book or resource to work out similar problems. I need to bone up on this aspect of trig.

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$\displaystyle\frac{\sec^2(\arctan(x)+\arctan(y))}{(x^2+1)}=\frac{\sec^2(\arctan(\frac{x+y}{1-xy}))}{(x^2+1)}=\frac{1+\tan^2(\arctan(\frac{x+y}{1-xy}))}{1+x^2}=\frac{1+(\frac{x+y}{1-xy})^2}{1+x^2}=\frac{y^2+1}{(1-xy)^2}$

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Thanks Abhra, very nice, appreciate it. There are quite a few things going on in this answer that I wish I knew the identities for. Even the first step in your transformation I am embarrassed to say I don't know. The remaining steps I fully understand. Any suggestions on a good resource that I could study? –  Joe Feb 10 '13 at 7:37
I guess I should rephrase my comment: What is the identity or property of arctan(x) + arctan(y) that allows it to be transformed to arctan((x+y)/(1-xy))? I don't know that one. –  Joe Feb 10 '13 at 7:50
Well, I see from Wikipedia it is a standard arctan addition identity I didn't know. Shame on me. I really need to memorize more identities. Thanks again. –  Joe Feb 10 '13 at 8:01
Joe, it's the arctan version of the identity $\tan(a+b)=(\tan a+\tan b)/(1-\tan a\tan b)$. –  Gerry Myerson Feb 10 '13 at 9:22