Inverse Trigonometric Functions Question 108
Question: $ 4{{\tan }^{-1}}\frac{1}{5}-{{\tan }^{-1}}\frac{1}{70}+{{\tan }^{-1}}\frac{1}{99}= $
[Roorkee 1981]
Options:
A) $ \frac{\pi }{2} $
B) $ \frac{\pi }{3} $
C) $ \frac{\pi }{4} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ 4{{\tan }^{-1}}\frac{1}{5}-{{\tan }^{-1}}\frac{1}{70}+{{\tan }^{-1}}\frac{1}{99} $
$ =2{{\tan }^{-1}}[ \frac{\frac{2}{5}}{1-\frac{1}{25}} ]-{{\tan }^{-1}}\frac{1}{70}+{{\tan }^{-1}}\frac{1}{99} $
$ =2{{\tan }^{-1}}( \frac{5}{12} )-{{\tan }^{-1}}\frac{1}{70}+{{\tan }^{-1}}\frac{1}{99} $
$ ={{\tan }^{-1}}[ \frac{\frac{5}{6}}{1-\frac{25}{144}} ]-{{\tan }^{-1}}\frac{1}{70}+{{\tan }^{-1}}\frac{1}{99} $
$ ={{\tan }^{-1}}( \frac{120}{119} )-{{\tan }^{-1}}\frac{1}{70}+{{\tan }^{-1}}\frac{1}{99} $
$ ={{\tan }^{-1}}( \frac{120}{119} )+{{\tan }^{-1}}[ \frac{\frac{1}{99}-\frac{1}{70}}{1+\frac{1}{99}.\frac{1}{70}} ] $
$ ={{\tan }^{-1}}( \frac{120}{119} )+{{\tan }^{-1}}( \frac{-29}{6931} ) $
$ ={{\tan }^{-1}}\frac{120}{119}-{{\tan }^{-1}}\frac{29}{6931}={{\tan }^{-1}}\frac{120}{119}-{{\tan }^{-1}}\frac{1}{239} $
$ ={{\tan }^{-1}}[ \frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119}\times \frac{1}{239}} ]={{\tan }^{-1}}(1)=\frac{\pi }{4} $ .
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