A) \[m+n\]
B) \[{{m}^{2}}-{{n}^{2}}\]
C) \[{{m}^{2}}+{{n}^{2}}\]
D) None of these
Correct Answer: C
Solution :
Given that \[a\,\cos \theta +b\,\sin \theta =m\] and \[a\,\sin \,\theta -b\,\cos \theta =n.\] Squaring and adding, we get \[{{(a\,\,\cos \theta +b\,\sin \theta )}^{2}}+{{(a\,\sin \theta -b\,\cos \theta )}^{2}}={{m}^{2}}+{{n}^{2}}\] \[\Rightarrow \,\,{{a}^{2}}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )+{{b}^{2}}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )\]\[+2ab\,(\cos \theta \,\sin \theta -\sin \theta \,\cos \theta )={{m}^{2}}+{{n}^{2}}\] Hence, \[{{a}^{2}}+{{b}^{2}}={{m}^{2}}+{{n}^{2}}.\] Trick: Here we can guess that the value of \[{{a}^{2}}+{{b}^{2}}\] is independent of q, so put any suitable value of q i.e. \[\frac{\pi }{2},\] so that \[b=m\] and \[a=n.\] Hence \[{{a}^{2}}+{{b}^{2}}={{m}^{2}}+{{n}^{2}}.\] (Also check for other value of q).
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