A) \[{{n}_{2}}>{{n}_{3}}>{{n}_{4}}>{{n}_{1}}\]
B) \[{{n}_{3}}>{{n}_{4}}>{{n}_{2}}>{{n}_{1}}\]
C) \[{{n}_{3}}={{n}_{1}}>{{n}_{4}}>{{n}_{2}}\]
D) \[{{n}_{2}}>{{n}_{1}}={{n}_{3}}>{{n}_{4}}\]
Correct Answer: D
Solution :
\[\because \sqrt{1+\sin 14}>1\Rightarrow {{n}_{3}}>{{n}_{4}}\] \[\sqrt{1+\sin 14}=\sqrt{{{(\sin 7+\cos 7)}^{2}}}\]\[=\sin 7+\cos 7\And \sqrt{\sin 7}>\sin 7\] \[\sqrt{\cos 7}>\cos 7\] \[\therefore {{n}_{2}}>{{n}_{1}}={{n}_{3}}>{{n}_{4}}\]You need to login to perform this action.
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