A) 10
B) \[{{2}^{10}}\]
C) \[{{2}^{9}}\]
D) 2
Correct Answer: D
Solution :
[d]: We have, \[\sin \theta +\cos ec\theta =2\] \[\Rightarrow \]\[{{\sin }^{2}}\theta +1=2\sin \theta \Rightarrow {{\sin }^{2}}\theta -2\sin \theta +1=0\] \[\Rightarrow \]\[{{(sin\theta -1)}^{2}}=0\Rightarrow \sin \theta =1\] \[\Rightarrow \]\[{{\sin }^{10}}\theta +\cos e{{c}^{10}}\theta ={{(1)}^{10}}+\frac{1}{{{(1)}^{10}}}=2\]
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