A) \[{{(mn)}^{-2/3}}\]
B) \[{{(mn)}^{2/3}}\]
C) \[{{(mn)}^{-1/3}}\]
D) \[{{(mn)}^{1/3}}\]
Correct Answer: A
Solution :
We have, \[mn=(\cos ec\,\theta -\sin \theta )\,(sec\theta -\cos \theta )\] \[=\left( \frac{1}{\sin \theta }-\sin \theta \right)\left( \frac{1}{\cos \theta }-\cos \theta \right)\] \[=\frac{1-{{\sin }^{2}}\theta }{\sin \theta }\times \frac{1-{{\cos }^{2}}\theta }{\cos \theta }\] \[=\frac{{{\cos }^{2}}\theta }{\sin \theta }\times \frac{{{\sin }^{2}}\theta }{\cos \theta }=\sin \theta .\cos \theta \] \[\therefore \,\,{{m}^{2/3}}+{{n}^{2/3}}={{\left( \frac{{{\cos }^{2}}\theta }{\sin \theta } \right)}^{2/3}}+{{\left( \frac{{{\sin }^{2}}\theta }{\cos \theta } \right)}^{2/3}}\] \[=\frac{{{\cos }^{4/3}}\theta }{{{\sin }^{2/3}}\theta }+\frac{{{\sin }^{4/3}}\theta }{{{\cos }^{2/3}}\theta }=\frac{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }{{{(\sin \theta .\cos \theta )}^{2/3}}}\] \[=\frac{1}{{{(mn)}^{2/3}}}={{(mn)}^{-2/3}}\]
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