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

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(C) Given that $a \cos 	heta + b \sin 	heta = m$ $(1)$ and $a \sin 	heta - b \cos 	heta = n$ $(2)$ Squaring both equations and adding them,we get:

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

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(C) Given that $a \cos 	heta + b \sin 	heta = m$ $(1)$ and $a \sin 	heta - b \cos 	heta = n$ $(2)$ 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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