A) 1
B) 2
C) \[2\sin \theta \]
D) \[sin\theta \,\cos \theta \]
Correct Answer: A
Solution :
We have, \[{{l}^{2}}{{m}^{2}}({{l}^{2}}+{{m}^{2}}+3)\] \[={{(\text{cosec }\theta \text{-sin}\theta \text{)}}^{2}}\,{{(\sec \theta -\cos \theta )}^{2}}\] \[\{{{(\cos ec\theta -\sin \theta )}^{2}}+{{(\sec \theta -\cos \theta )}^{2}}+3\}\] \[={{\left( \frac{1}{\sin \theta }-\sin \theta \right)}^{2}}{{\left( \frac{1}{\cos \theta }-\cos \theta \right)}^{2}}\] \[\left\{ {{\left( \frac{1-{{\sin }^{2}}\theta }{\sin \theta } \right)}^{2}}+{{\left( \frac{1-{{\cos }^{2}}\theta }{\cos \theta } \right)}^{2}}+3 \right\}\] \[=\frac{{{\cos }^{4}}\theta }{{{\sin }^{2}}\theta }\times \frac{{{\sin }^{4}}\theta }{{{\cos }^{2}}\theta }\,\,\left\{ \frac{{{\cos }^{4}}\theta }{{{\sin }^{2}}\theta }+\frac{{{\sin }^{4}}}{{{\cos }^{2}}\theta }+3 \right\}\] \[={{\cos }^{6}}\theta +{{\sin }^{6}}\theta +3{{\cos }^{2}}\theta {{\sin }^{2}}\theta \] \[=\{{{({{\cos }^{2}}\theta )}^{3}}+{{({{\sin }^{2}}\theta )}^{3}}+3{{\cos }^{2}}{{\sin }^{2}}\theta \] \[=\{{{({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )}^{3}}-3{{\cos }^{2}}\theta {{\sin }^{2}}\theta \] \[({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )\}+3{{\sin }^{2}}\theta {{\cos }^{2}}\theta \] \[=\{1-3{{\cos }^{2}}{{\sin }^{2}}\theta \}+3{{\cos }^{2}}\theta {{\sin }^{2}}\theta =1\]
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