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(A) Given $x = 3(\cos t + \sin t)$ and $y = 4(\cos t - \sin t)$.Squaring both equations:$x^2 = 9(\cos^2 t + \sin^2 t + 2 \sin t \cos t) = 9(1 + \sin 2

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$\frac{\sqrt{7}}{4}$

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(A) Given $x = 3(\cos t + \sin t)$ and $y = 4(\cos t - \sin t)$.Squaring both equations:$x^2 = 9(\cos^2 t + \sin^2 t + 2 \sin t \cos t) = 9(1 + \sin 2t)$$y^2 = 16(\cos^2 t + \sin^2 t - 2 \sin t \cos t) = 16(1 - \sin 2t)$From the first equation,$\sin 2t = \frac{x^2}{9} - 1$.Substituting this into the second equation:$y^2 = 16(1 - (\frac{x^2}{9} - 1)) = 16(2 - \frac{x^2}{9}) = 32 - \frac{16x^2}{9}$.Rearranging gives $\frac{16x^2}{9} + y^2 = 32$,or $\frac{x^2}{18} + \frac{y^2}{32} = 1$.This is an ellipse with $a^2 = 32$ and $b^2 = 18$.The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{18}{32}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$.

The eccentricity of the curve represented by $x = 3(\cos t + \sin t)$ and $y = 4(\cos t - \sin t)$ is

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