A) \[{{\tan }^{-1}}\left( 11 \right)\]
B) \[{{\tan }^{-1}}3\]
C) \[{{\tan }^{-1}}\left( \frac{2}{11} \right)\]
D) \[{{\tan }^{-1}}\left( \frac{11}{2} \right)\]
Correct Answer: D
Solution :
Also \[\tan \,3\alpha =\frac{3\tan \alpha -{{\tan }^{3}}\alpha }{1-3{{\tan }^{2}}\alpha }\] \[=\frac{3\times \frac{1}{2}-{{\left( \frac{1}{2} \right)}^{3}}}{1-3{{\left( \frac{1}{2} \right)}^{2}}}=\frac{\frac{3}{2}-\frac{1}{8}}{1-\frac{3}{4}}\] \[=\frac{\frac{11}{8}}{\frac{1}{4}}=\frac{11}{2}\Rightarrow \,\,3\alpha ={{\tan }^{-1}}\left( \frac{11}{2} \right)\]
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