A) \[m=1,n=1\]
B) \[m=1,n=4\]
C) \[m=4,n=1\]
D) \[m=1,n=-1\]
Correct Answer: B
Solution :
[b] \[{{p}^{2}}_{1}=4{{a}^{2}}{{\cos }^{2}}4\theta \] \[{{p}^{2}}_{2}=\frac{16{{a}^{2}}{{\cos }^{2}}2\theta }{{{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta }=16{{a}^{2}}{{\cos }^{2}}2\theta {{\cos }^{2}}\theta {{\sin }^{2}}\theta \]\[={{a}^{2}}{{\sin }^{2}}4\theta \] \[\therefore {{p}^{2}}_{1}+4{{p}^{2}}_{2}=4{{a}^{2}}\]You need to login to perform this action.
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