If (a+sqrt{2} b cos x)(a-sqrt{2} b cos y)=a^{ | vedclass

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(B) Given equation: $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^{2}-b^{2}$Expanding the left side:$a^{2} - \sqrt{2}ab \cos y + \sqrt{2}ab \cos x - 2

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$\frac{a+b}{a-b}$

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(B) Given equation: $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^{2}-b^{2}$Expanding the left side:$a^{2} - \sqrt{2}ab \cos y + \sqrt{2}ab \cos x - 2b^{2} \cos x \cos y = a^{2} - b^{2}$Subtracting $a^{2}$ from both sides:$-\sqrt{2}ab \cos y + \sqrt{2}ab \cos x - 2b^{2} \cos x \cos y = -b^{2}$Divide by $-1$:$\sqrt{2}ab \cos y - \sqrt{2}ab \cos x + 2b^{2} \cos x \cos y = b^{2}$Differentiating both sides with respect to $y$:$\sqrt{2}ab(-\sin y) - \sqrt{2}ab(-\sin x \frac{dx}{dy}) + 2b^{2}[\cos x(-\sin y) + \cos y(-\sin x \frac{dx}{dy})] = 0$At point $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$,$\sin x = \sin y = \frac{1}{\sqrt{2}}$ and $\cos x = \cos y = \frac{1}{\sqrt{2}}$:$-\sqrt{2}ab(\frac{1}{\sqrt{2}}) + \sqrt{2}ab(\frac{1}{\sqrt{2}} \frac{dx}{dy}) + 2b^{2}[-\frac{1}{2} - \frac{1}{2} \frac{dx}{dy}] = 0$$-ab + ab \frac{dx}{dy} - b^{2} - b^{2} \frac{dx}{dy} = 0$$\frac{dx}{dy}(ab - b^{2}) = ab + b^{2}$$\frac{dx}{dy} = \frac{ab + b^{2}}{ab - b^{2}} = \frac{b(a+b)}{b(a-b)} = \frac{a+b}{a-b}$

If $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^{2}-b^{2}$ where $a>b>0,$ then $\frac{dx}{dy}$ at $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$ is

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