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If acosÐ¤+bsinÐ¤=m and asinÐ¤-bcosÐ¤=nprove that, mÂ²+nÂ²=aÂ²+bÂ²

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$a \cos \phi + b\sin \phi =m \ \ \ \ \text{ Square on both side}\\ (a \cos \phi + b\sin \phi)^2 =m^2 \\ a^2 \cos^2 \phi + b^2 \sin^2 \phi + 2ab \sin \phi \cos \phi =m^2 ...(i)\\\\ a \sin \phi - b\cos \phi =n \ \ \ \ \text{ Square on both side}\\ (a \sin \phi - b\cos \phi)^2 =n^2 \\ a^2 \sin^2 \phi + b^2 \cos^2 \phi - 2ab \sin \phi \cos \phi =n^2 ...(ii)\\ \text{Equation (i) + Equation (ii)}\\ a^2 (\sin^2 \phi + \cos^2 \phi) + b^2 (\sin^2 \phi + \cos^2 \phi)=m^2+n^2 \\ a^2 + b^2 = m^2 + n^2 \ \ \because \sin^2 \phi + \cos^2 \phi =1 \\ \text{Hence Proved }$

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Ravindra Pindel

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