If (a + 2 bcosx ) ( a - 2 bcosy) = a ^ 2 - b ^ 2 , where a > b > 0,...

\left(a + \sqrt{2}bcosx\right)\left(a - \sqrt{2}bcosy\right) = {a}^{2} - {b}^{2}$</p><p>\left(a - \sqrt{2}bcosy\right)\left(- \sqrt{2}bsinx\right)\dfrac{dx}{dy} + \sqrt{2}bsiny\left(a + \sqrt{2}bcosx\right) = 0$

$\dfrac{dx}{dy} = \dfrac{\sqrt{2}bsiny\left(a + \sqrt{2}bcosx\right)}{\sqrt{2}bsinx\left(a - \sqrt{2}bcosy\right)} = \dfrac{\sin y\left(a + \sqrt{2}bcosx\right)}{\sin x\left(a - \sqrt{2}bcosy\right)}$

\dfrac{dx}{dy}\left|{}_{\left(\dfrac{\pi}{4}, \dfrac{\pi}{4}\right)} = \dfrac{a + b}{a - b}\right).