A) \[-2\]
B) \[-1\]
C) \[2/3\]
D) \[2\]
Correct Answer: A
Solution :
[a] \[\theta ={{\tan }^{-1}}\left[ \frac{2{{\tan }^{2}}\theta -\frac{1}{3}\tan \theta }{1+\frac{2}{3}{{\tan }^{3}}\theta } \right]\] \[\Rightarrow \tan \theta =\frac{6{{\tan }^{2}}\theta -\tan \theta }{3+2{{\tan }^{3}}\theta }\Rightarrow 1=\frac{6\tan \theta -1}{3+2{{\tan }^{3}}\theta }\] or \[\tan \theta =0\] \[\Rightarrow 2{{\tan }^{3}}\theta -6\tan \theta +4=0\] \[\Rightarrow {{(tan\theta -1)}^{2}}(tan\theta +2)=0\] \[\Rightarrow \tan \theta =1;\tan \theta =-2;\tan \theta =0.\]
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