A) \[mn\]
B) \[\tan \theta \]
C) \[1\]
D) \[\frac{m+n}{4}\]
Correct Answer: A
Solution :
Given, \[\cot \,\theta (1+\sin \theta )=4m\] ?(i) and \[\cot \theta (1-\sin \theta )=4n\] ?.(ii) From Eq. \[{{(i)}^{2}}-\] Eq. \[{{(ii)}^{2}}\], we get \[{{(4m)}^{2}}-{{(4n)}^{2}}={{\cot }^{2}}\theta {{(1+\sin \theta )}^{2}}\] \[-{{\cot }^{2}}\theta {{(1-\sin \,\theta )}^{2}}\] \[={{\cot }^{2}}\theta \,\{{{(1+\sin \theta )}^{2}}-{{(1-\sin \theta )}^{2}}\}\] \[={{\cot }^{2}}\theta \{(1+\sin \theta +1-\sin \theta )\] \[(1+\sin \theta -1+\sin \theta )\}\] \[={{\cot }^{2}}\theta \{2.2\,\sin \theta \}\] \[=4\frac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }\times \sin \theta =4\frac{{{\cos }^{2}}\theta }{\sin \theta }\] \[\Rightarrow \] \[{{m}^{2}}-{{n}^{2}}=\frac{{{\cos }^{2}}\theta }{4\sin \theta }\] Squaring on both sides, we get \[({{m}^{2}}-{{n}^{2}})=\frac{{{\cos }^{4}}\theta }{16{{\sin }^{2}}\theta }\] ?.(iii) Now, on multiplying Eq. (i) by Eq. (ii), we get \[{{\cot }^{2}}\theta (1-{{\sin }^{2}}\theta )=16mn\] \[\Rightarrow \] \[mn=\frac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }\,(1-{{\sin }^{2}}\theta ).\frac{1}{16}\] \[\Rightarrow \] \[mn=\frac{{{\cos }^{4}}\theta }{16{{\sin }^{2}}\theta }\] ?.(iv) From Eqs. (iii) and (v), we get \[{{({{m}^{2}}-{{n}^{2}})}^{2}}=mn.\]
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