A) \[\frac{1}{2}{{\cos }^{-1}}\left( \frac{3}{5} \right)\]
B) \[\frac{1}{2}{{\sin }^{-1}}\left( \frac{3}{5} \right)\]
C) \[\frac{1}{2}{{\tan }^{-1}}\left( \frac{3}{5} \right)\]
D) \[{{\tan }^{-1}}\left( \frac{1}{2} \right)\]
Correct Answer: D
Solution :
\[{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right)\] \[\therefore {{\tan }^{-1}}\frac{1}{4}+{{\tan }^{-1}}\frac{2}{9}={{\tan }^{-1}}\left( \frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4}\times \frac{2}{9}} \right)\] \[\left( \because x=\frac{1}{4},y=\frac{2}{9} \right)\] \[={{\tan }^{-1}}\left( \frac{1}{2} \right)\]
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