# 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.

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