The locus of points (x, y) in the plane satis | vedclass

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(C) Given the equation $\sin ^2 x + \sin ^2 y = 1$. Using the identity $\sin ^2 y = 1 - \sin ^2 x = \cos ^2 x$. This implies $\sin y = \pm \cos x$. Ca

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infinitely many lines with slope $\pm 1$

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(C) Given the equation $\sin ^2 x + \sin ^2 y = 1$. Using the identity $\sin ^2 y = 1 - \sin ^2 x = \cos ^2 x$. This implies $\sin y = \pm \cos x$. Case $1$: $\sin y = \cos x = \sin(\frac{\pi}{2} - x)$. The general solution is $y = n\pi + (-1)^n(\frac{\pi}{2} - x)$,which represents lines with slope $\pm 1$. Case $2$: $\sin y = -\cos x = \sin(x - \frac{\pi}{2})$. The general solution is $y = n\pi + (-1)^n(x - \frac{\pi}{2})$,which also represents lines with slope $\pm 1$. Since $n$ can be any integer,there are infinitely many such lines.

The locus of points $(x, y)$ in the plane satisfying $\sin ^2 x + \sin ^2 y = 1$ consists of

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