If a cos theta + b sin theta = m and a sin th | vedclass

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(C) Given that $a \cos 	heta + b \sin 	heta = m$ and $a \sin 	heta - b \cos 	heta = n.$Squaring both equations and adding them,we get:${(a \cos

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${m^2} + {n^2}$

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(C) Given that $a \cos 	heta + b \sin 	heta = m$ and $a \sin 	heta - b \cos 	heta = n.$Squaring both equations and adding them,we get:${(a \cos 	heta + b \sin 	heta)^2} + {(a \sin 	heta - b \cos 	heta)^2} = {m^2} + {n^2}$Expanding the squares:$({a^2} \cos^2 	heta + {b^2} \sin^2 	heta + 2ab \sin 	heta \cos 	heta) + ({a^2} \sin^2 	heta + {b^2} \cos^2 	heta - 2ab \sin 	heta \cos 	heta) = {m^2} + {n^2}$Grouping the terms:${a^2}(\cos^2 	heta + \sin^2 	heta) + {b^2}(\sin^2 	heta + \cos^2 	heta) = {m^2} + {n^2}$Since $\sin^2 	heta + \cos^2 	heta = 1$,we have:${a^2}(1) + {b^2}(1) = {m^2} + {n^2}$Therefore,${a^2} + {b^2} = {m^2} + {n^2}.$

If $a \cos \theta + b \sin \theta = m$ and $a \sin \theta - b \cos \theta = n,$ then ${a^2} + {b^2} = $

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