A) \[{{m}^{2}}+{{n}^{2}}\]
B) \[{{m}^{2}}-{{n}^{2}}\]
C) \[mn\]
D) \[m+n\]
Correct Answer: A
Solution :
Given, \[a\,\cos \,\theta \,+b\,\sin \theta \,=m\] ...(i) : and \[a\,\cos \,\theta \,-b\,\sin \theta =n\] ...(ii) On squaring and adding Eqs. (i) and (ii), we get \[{{(a\,\cos \,\theta \,+b\,\sin \theta )}^{2}}+\,{{(b\,\cos \theta -a\,\sin \,\theta )}^{2}}={{m}^{2}}+{{n}^{2}}\] i.e., \[{{a}^{2}}\,{{\cos }^{2}}\,\theta +{{b}^{2}}\,{{\sin }^{2}}\theta +2ab\,\sin \,\theta \cdot \cos \,\theta +{{b}^{2}}\,{{\cos }^{2}}\,\theta \]\[+{{a}^{2}}\,{{\sin }^{2}}\,\theta \] \[-2ab\,\sin \,\theta \cdot \cos \theta ={{m}^{2}}+{{n}^{2}}\] \[({{a}^{2}}+{{b}^{2}})\,{{\cos }^{2}}\,\theta +\,({{a}^{2}}+{{b}^{2}})\,{{\sin }^{2}}\theta ={{m}^{2}}+{{n}^{2}}\] \[\Rightarrow \] \[({{a}^{2}}+{{b}^{2}})\,({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )\,={{m}^{2}}+{{n}^{2}}\] i.e., \[{{a}^{2}}+{{b}^{2}}={{m}^{2}}+{{n}^{2}}\]
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