A) 0
B) - 1
C) 1
D) 2
Correct Answer: D
Solution :
\[\text{cose}{{\text{c}}^{2}}\theta +{{\sin }^{2}}\theta =\frac{1}{{{\sin }^{2}}\theta }+{{\sin }^{2}}\theta \] \[={{\left( \frac{1}{\sin \theta }-\sin \theta \right)}^{2}}+2\] \[\therefore \] Minimum value of \[(\text{cose}{{\text{c}}^{2}}\theta +{{\sin }^{2}}\theta )=2\] [because the minimum value of \[{{\left( \frac{1}{\sin \theta }-\sin \theta \right)}^{2}}\]0]
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