A) \[{{\tan }^{-1}}\frac{1}{2}\]
B) \[{{\tan }^{-1}}(2)\]
C) \[{{\tan }^{-1}}\left( \frac{2}{3} \right)\]
D) \[{{\tan }^{-1}}\left( \frac{\sqrt{3}}{4} \right)\]
Correct Answer: B
Solution :
[b] \[\tan \,\theta '=\frac{\tan \,\theta }{\cos \,\alpha }=\frac{\tan {{60}^{o}}}{\cos \,{{30}^{o}}}=\frac{\sqrt{3}}{\sqrt{3}/2}=2\] \[\therefore \,\,\,\,\theta '=ta{{n}^{-1}}(2).\]
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