A) 1
B) 2
C) 3
D) 0
Correct Answer: B
Solution :
| [b] \[\because \]\[\sin P\text{cosec}\,P=2\] \[\Rightarrow \] \[\sin P+\frac{1}{\sin P}=2\]\[\Rightarrow \]\[{{\sin }^{2}}P+1=\sin P\] \[\Rightarrow \] \[{{\sin }^{2}}P+1-2\sin P=0\] \[\Rightarrow \] \[{{(\sin P-1)}^{2}}=0\] \[\Rightarrow \] \[\sin P-1=0\] \[\Rightarrow \] \[\sin P=1\] \[\therefore \] \[{{\sin }^{7}}P+\text{cose}{{\text{c}}^{7}}P={{\sin }^{7}}P+\frac{1}{{{\sin }^{2}}P}\] \[={{(\sin P)}^{7}}+\frac{1}{{{(\sin P)}^{7}}}={{(1)}^{7}}+\frac{1}{{{(1)}^{7}}}=1+\frac{1}{1}=1+1=2\] |
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