Inverse Trigonometric Functions Question 28
Question: $ 4{{\tan }^{-1}}\frac{1}{5}-{{\tan }^{-1}}\frac{1}{239} $ is equal to
[MNR 1995]
Options:
A) $ \pi $
B) $ \frac{\pi }{2} $
C) $ \frac{\pi }{3} $
D) $ \frac{\pi }{4} $
Show Answer
Answer:
Correct Answer: D
Solution:
Since $ 2{{\tan }^{-1}}x={{\tan }^{-1}}\frac{2x}{1-x^{2}} $ \ $ 4{{\tan }^{-1}}\frac{1}{5}=2[ 2{{\tan }^{-1}}\frac{1}{5} ]=2{{\tan }^{-1}}\frac{\frac{2}{5}}{1-\frac{1}{25}} $
$ =2{{\tan }^{-1}}\frac{10}{24}={{\tan }^{-1}}\frac{\frac{20}{24}}{1-\frac{100}{576}}={{\tan }^{-1}}\frac{120}{119} $ So, $ 4{{\tan }^{-1}}\frac{1}{5}-{{\tan }^{-1}}\frac{1}{239}={{\tan }^{-1}}\frac{120}{119}-{{\tan }^{-1}}\frac{1}{239} $
$ ={{\tan }^{-1}}\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119}.\frac{1}{239}}={{\tan }^{-1}}\frac{(120\times 239)-119}{(119\times 239)+120} $
Therefore $ {{\tan }^{-1}}1=\frac{\pi }{4} $ .
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