If theta is the acute angle between the curve | vedclass

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(A) Given curves are:$x^2+y^2=2020 \sqrt{2} \quad (i)$$x^2-y^2=2020 \quad (ii)$Adding $(i)$ and $(ii)$, we get:$2x^2 = 2020(\sqrt{2}+1) \implies x^2 =

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$\sqrt{2}$

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(A) Given curves are:$x^2+y^2=2020 \sqrt{2} \quad (i)$$x^2-y^2=2020 \quad (ii)$Adding $(i)$ and $(ii)$, we get:$2x^2 = 2020(\sqrt{2}+1) \implies x^2 = 1010(\sqrt{2}+1)$Subtracting $(ii)$ from $(i)$, we get:$2y^2 = 2020(\sqrt{2}-1) \implies y^2 = 1010(\sqrt{2}-1)$For curve $(i)$, differentiating with respect to $x$:$2x + 2yy' = 0 \implies y' = -\frac{x}{y} = m_1$For curve $(ii)$, differentiating with respect to $x$:$2x - 2yy' = 0 \implies y' = \frac{x}{y} = m_2$The angle $	heta$ between the curves is given by:$	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight| = \left| \frac{-\frac{x}{y} - \frac{x}{y}}{1 + (-\frac{x}{y})(\frac{x}{y})} ight| = \left| \frac{-\frac{2x}{y}}{1 - \frac{x^2}{y^2}} ight| = \left| \frac{-2xy}{y^2 - x^2} ight|$From $(ii)$, $y^2 - x^2 = -2020$. Also $x^2y^2 = (1010)^2(\sqrt{2}+1)(\sqrt{2}-1) = (1010)^2$, so $xy = \pm 1010$.$	an 	heta = \left| \frac{-2(\pm 1010)}{-2020} ight| = |\pm 1| = 1$Since $	heta$ is acute, $	heta = \frac{\pi}{4}$.Therefore, $\frac{\sin 	heta + \cos 	heta}{	an 	heta} = \frac{\sin(\pi/4) + \cos(\pi/4)}{	an(\pi/4)} = \frac{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}}{1} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

If $\theta$ is the acute angle between the curves $x^2+y^2=2020 \sqrt{2}$ and $x^2-y^2=2020$, then $\frac{\sin \theta+\cos \theta}{\tan \theta}$ is equal to

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