A) \[\frac{\pi }{6}\]
B) \[\frac{3\pi }{4}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{2}\]
Correct Answer: B
Solution :
[b] Given: \[{{x}^{2}}-{{y}^{2}}{{\sec }^{2}}\theta =4\] and \[{{x}^{2}}{{\sec }^{2}}\theta +{{y}^{2}}=16\] \[\Rightarrow \frac{{{x}^{2}}}{4}-\frac{{{y}^{2}}}{4{{\cos }^{2}}\theta }=1\] And \[\frac{{{x}^{2}}}{16{{\cos }^{2}}\theta }+\frac{{{y}^{2}}}{16}=1\] According to problem \[\frac{4+4{{\cos }^{2}}\theta }{4}=3\left( \frac{16-16{{\cos }^{2}}\theta }{16} \right)\] \[\Rightarrow 1+{{\cos }^{2}}\theta =3(1-co{{s}^{2}}\theta )\Rightarrow 4co{{s}^{2}}\theta =2\] \[\Rightarrow \cos \theta =\pm \frac{1}{\sqrt{2}}\Rightarrow \theta =\frac{\pi }{4},\frac{3\pi }{4}\]
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