A) \[\frac{12}{25}\]
B) \[\frac{24}{25}\]
C) \[\frac{1}{5}\]
D) None of these
Correct Answer: B
Solution :
\[\sin \left( 4{{\tan }^{-1}}\frac{1}{3} \right)=\sin \left[ 2{{\tan }^{-1}}\left( \frac{2/3}{1-(1/9)} \right) \right]\] \[=\sin \left[ 2{{\tan }^{-1}}\frac{3}{4} \right]=\sin {{\sin }^{-1}}\left( \frac{2\times (3/4)}{1+(9/16)} \right)\] \[=\frac{3}{2}\times \frac{16}{25}=\frac{24}{25}\] \[\left( \because 2{{\tan }^{-1}}x={{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}} \right)\]
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