A) \[k=1\]
B) \[k=-{{\sin }^{2}}\theta {{\cos }^{2}}\theta \]
C) \[k={{\sin }^{2}}\theta \]
D) \[k={{\cos }^{2}}\theta \]
Correct Answer: B
Solution :
\[{{p}_{n}}-{{p}_{n-2}}=({{\cos }^{n}}\theta +{{\sin }^{n}}\theta )-({{\cos }^{n-2}}\theta +{{\sin }^{n-2}}\theta )\] \[={{\cos }^{n-2}}\theta ({{\cos }^{2}}\theta -1)+{{\sin }^{n-2}}\theta ({{\sin }^{2}}\theta -1)\] \[=-{{\sin }^{2}}\theta {{\cos }^{n-2}}\theta -{{\cos }^{2}}\theta {{\sin }^{n-2}}\theta \] \[=-{{\sin }^{2}}\theta {{\cos }^{2}}\theta ({{\cos }^{n-4}}\theta +{{\sin }^{n-4}}\theta )\] \[={{\sin }^{2}}\theta {{\cos }^{2}}\theta {{p}_{n-4}}=k{{p}_{n-4}}\] \[\Rightarrow \,\,k=-{{\sin }^{2}}\theta {{\cos }^{2}}\theta \]
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