A) 1
B) 4
C) 2
D) none of these
Correct Answer: C
Solution :
Since, \[\sin \theta +\cos ec\,\theta =2\] \[\Rightarrow \] \[{{(sin\theta +cosec\theta )}^{2}}={{2}^{2}}\] \[\Rightarrow \]\[{{\sin }^{2}}\theta +\cos e{{c}^{2}}\theta +2\sin \theta \cos ec\theta =4\] \[\Rightarrow \]\[{{\sin }^{2}}\theta +\cos e{{c}^{2}}\theta +2=4\] \[\Rightarrow \]\[{{\sin }^{2}}\theta +\cos e{{c}^{2}}\theta =2\]
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