A) \[{{m}^{2}}-{{n}^{2}}\]
B) \[{{m}^{2}}+{{n}^{2}}\]
C) \[{{m}^{2}}{{n}^{2}}\]
D) \[{{n}^{2}}-{{m}^{2}}\]
Correct Answer: B
Solution :
\[a\cos \theta +b\sin \theta =m\] ? (i) \[a\sin \theta -b\cos \theta =n\] ? (ii) Squaring and adding, \[{{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta +2ab\sin \theta \]\[\cos \theta +\theta +{{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta -\]\[2ab\sin \theta .\cos \theta ={{m}^{2}}+{{n}^{2}}\] \[\Rightarrow \]\[{{a}^{2}}{{\cos }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta \]\[={{m}^{2}}+{{n}^{2}}\] \[\Rightarrow \]\[{{a}^{2}}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )+{{b}^{2}}({{\sin }^{2}}\theta +{{\cos }^{2}})\]\[={{m}^{2}}+{{n}^{2}}\] \[\Rightarrow \]\[{{a}^{2}}+{{b}^{2}}={{m}^{2}}+{{n}^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
