| If \[0\le \theta \le \frac{\pi }{2}\] and \[{{\sec }^{2}}\theta +{{\tan }^{2}}\theta =7,\] then \[\theta \] is |
A) \[\frac{5\pi }{12}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{\pi }{5}\]
D) \[\frac{\pi }{6}\]
Correct Answer: B
Solution :
| Given, \[{{\sec }^{2}}\theta +{{\tan }^{2}}\theta =7\] |
| \[\Rightarrow \] \[1+{{\tan }^{2}}\theta +{{\tan }^{2}}\theta =7\] \[[\because {{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta ]\] |
| \[\Rightarrow \] \[1+2{{\tan }^{2}}\theta =7\] |
| \[\Rightarrow \] \[2{{\tan }^{2}}\theta =6\]\[\Rightarrow \]\[{{\tan }^{2}}\theta =3\] |
| \[\therefore \] \[\tan \theta =\pm \,\sqrt{3}\] |
| Since, \[0\le \theta \le \frac{\pi }{2}\] |
| \[\therefore \] \[\tan \theta =\sqrt{3}\] |
| \[\theta =60{}^\circ =\frac{\pi }{3}\] [as \[\pi =180{}^\circ \]] |
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