A) \[{{m}^{2}}-{{n}^{2}}=4mn\]
B) \[{{m}^{2}}+{{n}^{2}}=4mn\]
C) \[{{m}^{2}}-{{n}^{2}}={{m}^{2}}+{{n}^{2}}\]
D) \[{{m}^{2}}-{{n}^{2}}=4\sqrt{mn}\]
Correct Answer: D
Solution :
\[(m+n)=2\,\tan \theta ,\,\,m-n=2\,\sin \theta \] \[\therefore \,\,\,{{m}^{2}}-{{n}^{2}}=4\,\tan \theta \,.\,\sin \theta \] ?..(i) \[4\sqrt{mn}=4\sqrt{{{\tan }^{2}}\theta -{{\sin }^{2}}\theta }=4\,\sin \theta \,.\,\tan \theta \] ?..(ii) From (i) and (ii), \[{{m}^{2}}-{{n}^{2}}=4\sqrt{mn}\].
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