A) 2
B) \[{{2}^{n}}\]
C) \[{{2}^{n-1}}\]
D) \[{{2}^{n-2}}\]
Correct Answer: A
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, \[{{\sin }^{n}}x+\cos e{{c}^{n}}x={{(1)}^{n}}+{{(1)}^{n}}\] \[=1+1=2\]
You need to login to perform this action.
You will be redirected in
3 sec
