A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{2}\]
C) \[\pi \]
D) None of these
Correct Answer: C
Solution :
[c] we have, \[3{{\tan }^{-1}}\frac{1}{2}+2{{\tan }^{-1}}\frac{1}{5}+{{\sin }^{-1}}\frac{142}{65\sqrt{5}}\] \[=2\left( {{\tan }^{-1}}\frac{1}{2}+{{\tan }^{-1}}\frac{1}{5} \right)+{{\tan }^{-1}}\frac{1}{2}+{{\tan }^{-1}}\frac{142}{31}\] \[=2{{\tan }^{-1}}\frac{7}{9}+\pi +{{\tan }^{-1}}\frac{\frac{1}{2}+\frac{142}{31}}{1-\left( \frac{1}{2} \right)\left( \frac{142}{31} \right)}\] \[={{\tan }^{-1}}\frac{\frac{14}{9}}{1-\frac{49}{81}}+\pi -{{\tan }^{-1}}\frac{315}{80}\] \[=\pi -{{\tan }^{-1}}\frac{63}{16}+{{\tan }^{-1}}\frac{63}{16}=\pi \]
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