A) 2
B) \[{{2}^{n}}\]
C) \[{{2}^{n-1}}\]
D) \[{{2}^{n-2}}\]
Correct Answer: A
Solution :
\[\sin x+\cos \text{ec}x=2\Rightarrow {{(\sin x-1)}^{2}}=0\Rightarrow \sin x=1\] \[\therefore {{\sin }^{n}}x+\text{cose}{{\text{c}}^{n}}x=1+1=2\] .
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