A) \[\frac{{{x}^{2}}}{3}+\frac{{{y}^{2}}}{4}=1\]
B) \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{3}=1\]
C) \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{3}=\frac{4}{3}\]
D) None of these
Correct Answer: B
Solution :
\[sin\text{ }x+cosec\text{ }x=2\] \[\sin x+\frac{1}{\sin x}=2\] \[{{\sin }^{2}}x+1=2\sin x\] \[\Rightarrow \] \[si{{n}^{2}}x+1=2\text{ }sin\text{ }x\] \[\Rightarrow \] \[{{(\sin x-1)}^{2}}=0\] \[\Rightarrow \] \[\sin x=1\] \[\Rightarrow \] \[\cos ec\,x=1\] Now, \[{{\sin }^{n}}x+\cos e{{c}^{n}}x={{(1)}^{n}}+{{(1)}^{n}}\] \[=1+1=2\]
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