If x sin^3 alpha + y cos^3 alpha = sin alpha | vedclass

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(B) Given equations are:$x \sin^3 \alpha + y \cos^3 \alpha = \sin \alpha \cos \alpha$ ... $(i)$$x \sin \alpha - y \cos \alpha = 0$ ... $(ii)$From $(ii

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(B) Given equations are:$x \sin^3 \alpha + y \cos^3 \alpha = \sin \alpha \cos \alpha$ ... $(i)$$x \sin \alpha - y \cos \alpha = 0$ ... $(ii)$From $(ii)$,we have $x \sin \alpha = y \cos \alpha$.Substitute this into $(i)$:$(x \sin \alpha) \sin^2 \alpha + y \cos^3 \alpha = \sin \alpha \cos \alpha$$(y \cos \alpha) \sin^2 \alpha + y \cos^3 \alpha = \sin \alpha \cos \alpha$$y \cos \alpha (\sin^2 \alpha + \cos^2 \alpha) = \sin \alpha \cos \alpha$Since $\sin^2 \alpha + \cos^2 \alpha = 1$,we get:$y \cos \alpha = \sin \alpha \cos \alpha$Assuming $\cos \alpha 
eq 0$,we get $y = \sin \alpha$.Substituting $y = \sin \alpha$ into $x \sin \alpha = y \cos \alpha$,we get $x \sin \alpha = \sin \alpha \cos \alpha$,which implies $x = \cos \alpha$.Therefore,$x^2 + y^2 = \cos^2 \alpha + \sin^2 \alpha = 1$.

If $x \sin^3 \alpha + y \cos^3 \alpha = \sin \alpha \cos \alpha$ and $x \sin \alpha - y \cos \alpha = 0$,then $x^2 + y^2 = $

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