A) \[15{}^\circ \] or\[\text{ }0{}^\circ \]
B) \[30{}^\circ \]or \[0{}^\circ \]
C) \[45{}^\circ ~\]or \[0{}^\circ \]
D) \[60{}^\circ \]or\[0{}^\circ \]
Correct Answer: D
Solution :
| [d] \[{{\tan }^{2}}\theta +3=3\sec \theta \] \[{{\tan }^{2}}\theta +3=3\sqrt{1+{{\tan }^{2}}\theta }\] Squaring on both sides, we get \[{{(ta{{n}^{2}}\theta +3)}^{2}}=9\,(1+ta{{n}^{2}}\theta )\] \[{{\tan }^{4}}\theta +9+6{{\tan }^{2}}\theta =9+9{{\tan }^{2}}\theta {{\tan }^{4}}\theta =3{{\tan }^{2}}\theta \] \[{{\tan }^{2}}\theta \,(ta{{n}^{2}}\theta -3)=0\] \[\tan \theta =\sqrt{3}\]or \[\tan \theta =0\] \[\theta =60{}^\circ \]or \[\theta -0{}^\circ \] |
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