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The locus of points \((x, y)\) in the plane satisfying \(\sin ^2 x+\sin ^2 y=1\) consists of
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Verified Answer
The correct answer is:
infinitely many lines with slope \(\pm 1\)
\(\begin{array}{r}\text {Hint: }
\sin ^2 y=\cos ^2 x \\
\sin y= \pm \cos x
\end{array}\)
\(\begin{aligned}
& \text { If } \sin y=\cos x=\sin \left(\frac{\pi}{2}-x\right) \\
& \Rightarrow y=n \pi+(-1)^n\left(\frac{\pi}{2}-x\right)
\end{aligned}\)
\(\text { If } \sin y=-\cos x=\sin \left(x-\frac{\pi}{2}\right)\)
\(\Rightarrow \mathrm{y}=\mathrm{n} \pi+(-1)^{\mathrm{n}}\left(\mathrm{x}-\frac{\pi}{2}\right)\)
\sin ^2 y=\cos ^2 x \\
\sin y= \pm \cos x
\end{array}\)
\(\begin{aligned}
& \text { If } \sin y=\cos x=\sin \left(\frac{\pi}{2}-x\right) \\
& \Rightarrow y=n \pi+(-1)^n\left(\frac{\pi}{2}-x\right)
\end{aligned}\)
\(\text { If } \sin y=-\cos x=\sin \left(x-\frac{\pi}{2}\right)\)
\(\Rightarrow \mathrm{y}=\mathrm{n} \pi+(-1)^{\mathrm{n}}\left(\mathrm{x}-\frac{\pi}{2}\right)\)
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