A) \[\frac{\pi }{3}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{2}\]
D) \[\pi \]
E) \[2\pi \]
Correct Answer: C
Solution :
\[{{\sin }^{-1}}\left( \frac{4}{5} \right)+2{{\tan }^{-1}}\left( \frac{1}{3} \right)\] \[={{\tan }^{-1}}\left( \frac{4/5}{\sqrt{1-{{\left( \frac{4}{5} \right)}^{2}}}} \right)+{{\tan }^{-1}}\left( \frac{2.\frac{1}{3}}{1-{{\left( \frac{1}{3} \right)}^{2}}} \right)\] \[\left( \begin{align} & \because {{\sin }^{-1}}x={{\tan }^{-1}}\frac{x}{\sqrt{1-{{x}^{2}}}}and \\ & 2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)if\,-1<x<1 \\ \end{align} \right)\] \[={{\tan }^{-1}}\frac{4}{3}+{{\tan }^{-1}}\left( \frac{3}{4} \right)\] \[={{\tan }^{-1}}\left( \frac{\frac{4}{3}+\frac{3}{4}}{1-\frac{4}{3}.\frac{3}{4}} \right)={{\tan }^{-1}}(\infty )\] \[=\frac{\pi }{2}\]
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