arctan((x+y)/(1-xy)) just do it in respect to x and I can do the rest.
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Originally Posted by daskywalker arctan((x+y)/(1-xy)) just do it in respect to x and I can do the rest. Tour de-force $\displaystyle \frac{1}{1 +\left( \frac{x+y}{1-xy}\right)^2} \cdot \frac{1+y^2}{(1-xy)^2} = \frac{1}{1+x^2} $
Originally Posted by daskywalker arctan((x+y)/(1-xy)) just do it in respect to x and I can do the rest. Or using this identity that makes it easier $\displaystyle \tan ^{-1} \frac{x+y}{1-xy} = \tan ^{-1} x + \tan ^{-1} y. $
Originally Posted by Danny Tour de-force $\displaystyle \frac{1}{1 +\left( \frac{x+y}{1-xy}\right)^2} \cdot \frac{1+y^2}{(1-xy)^2} = \frac{1}{1+x^2} $ forgive my stupidity but I don't exactly see the simplification.I got to the point where (y^2+1)/{y^2(1+x^2)+(1+x^2)}
Originally Posted by daskywalker forgive my stupidity but I don't exactly see the simplification.I got to the point where (y^2+1)/{y^2(1+x^2)+(1+x^2)} When you multiply the denominator through by $\displaystyle (1-xy)^2$ you get $\displaystyle (1-xy)^2+(x+y)^2 = 1 - 2xy + x^2 y^2 + x^2 + 2xy +y^2 = (1+x^2)(1+y^2). $
Originally Posted by Danny When you multiply the denominator through by $\displaystyle (1-xy)^2$ you get $\displaystyle (1-xy)^2+(x+y)^2 = 1 - 2xy + x^2 y^2 + x^2 + 2xy +y^2 = (1+x^2)(1+y^2). $ ok got it now. many thanks!
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