If x = a cos theta + b sin theta and y = b co | vedclass

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(B) Given:$x = a \cos 	heta + b \sin 	heta$$y = b \cos 	heta - a \sin 	heta$Squaring both equations:$x^{2} = (a \cos 	heta + b \sin 	heta)^{2} =

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$a^{2} + b^{2}$

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(B) Given:$x = a \cos 	heta + b \sin 	heta$$y = b \cos 	heta - a \sin 	heta$Squaring both equations:$x^{2} = (a \cos 	heta + b \sin 	heta)^{2} = a^{2} \cos^{2} 	heta + b^{2} \sin^{2} 	heta + 2ab \sin 	heta \cos 	heta$$y^{2} = (b \cos 	heta - a \sin 	heta)^{2} = b^{2} \cos^{2} 	heta + a^{2} \sin^{2} 	heta - 2ab \cos 	heta \sin 	heta$Adding $x^{2}$ and $y^{2}$:$x^{2} + y^{2} = (a^{2} \cos^{2} 	heta + b^{2} \sin^{2} 	heta + 2ab \sin 	heta \cos 	heta) + (b^{2} \cos^{2} 	heta + a^{2} \sin^{2} 	heta - 2ab \cos 	heta \sin 	heta)$$x^{2} + y^{2} = a^{2} (\cos^{2} 	heta + \sin^{2} 	heta) + b^{2} (\sin^{2} 	heta + \cos^{2} 	heta)$Since $\sin^{2} 	heta + \cos^{2} 	heta = 1$,we get:$x^{2} + y^{2} = a^{2}(1) + b^{2}(1) = a^{2} + b^{2}$

If $x = a \cos \theta + b \sin \theta$ and $y = b \cos \theta - a \sin \theta,$ then $x^{2} + y^{2}$ is equal to

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