A) \[{{m}^{2}}-{{n}^{2}}\]
B) \[{{n}^{2}}-{{m}^{2}}\]
C) \[{{m}^{2}}+{{n}^{2}}\]
D) \[{{m}^{2}}{{n}^{2}}\]
Correct Answer: C
Solution :
| [c] We have,\[a\cos \theta +b\sin \theta =m\]...(1) |
| \[a\sin \theta -b\cos \theta =n\] ...(2) |
| Squaring and adding eqs. (1) and (2), we get |
| \[{{m}^{2}}+{{n}^{2}}=({{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta +2ab\cos \theta \sin \theta )\]\[+({{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta -2ab\sin \theta \cos \theta )\] |
| \[={{a}^{2}}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )+{{b}^{2}}({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )\] |
| \[={{a}^{2}}+{{b}^{2}}\]\[[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1]\] |
| Hence, \[{{a}^{2}}+{{b}^{2}}={{m}^{2}}+{{n}^{2}}\] |
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