A) \[\frac{8}{\sqrt{7}}\]
B) \[\frac{16}{7}\]
C) \[\frac{4}{\sqrt{7}}\]
D) \[\frac{8}{7}\]
Correct Answer: A
Solution :
\[\cos 2Q=1/7\] \[=2{{\cos }^{2}}Q-1=1/7\] \[=2{{\cos }^{2}}Q=8/7\] \[{{\cos }^{2}}Q=4/7\] \[=\frac{c{{p}^{2}}}{4}=\frac{4}{7}\] \[=Cp=\frac{4}{\sqrt{7}}\] \[\sec 2Q=7\] \[=\frac{2}{2{{\cos }^{2}}Q-1}=7\] \[=2{{\left( \frac{C{{p}_{2}}}{2} \right)}^{2}}-1=\frac{1}{7}\] \[=2{{\left( \frac{C{{p}_{2}}}{2} \right)}^{2}}=\frac{8}{7}\] \[=\]\[\frac{4}{\sqrt{7}}+\frac{4}{\sqrt{7}}=\frac{8}{\sqrt{7}}\]You need to login to perform this action.
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