Inverse Trigonometric Functions Question 125
Question: The value of $ 3{{\tan }^{-1}}\frac{1}{2}+2{{\tan }^{-1}}\frac{1}{5}+{{\sin }^{-1}}\frac{142}{65\sqrt{5}} $ is
Options:
A) $ \frac{\pi }{4} $
B) $ \frac{\pi }{2} $
C) $ \pi $
D) None of these
Show Answer
Answer:
Correct Answer: C
we have, $ 3{{\tan }^{-1}}\frac{1}{2}+2{{\tan }^{-1}}\frac{1}{5}+{{\sin }^{-1}}\frac{142}{65\sqrt{5}} $
$ =2( {{\tan }^{-1}}\frac{1}{2}+{{\tan }^{-1}}\frac{1}{5} )+{{\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-( \frac{1}{2} )( \frac{142}{31} )} $
$ ={{\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 $
BETA
