Inverse Trigonometric Functions Question 136
Question: $ {{\tan }^{-1}}( \frac{1}{4} )+{{\tan }^{-1}}( \frac{2}{9} )= $
[EAMCET 1994]
Options:
A) $ \frac{1}{2}{{\cos }^{-1}}( \frac{3}{5} ) $
B) $ \frac{1}{2}{{\sin }^{-1}}( \frac{3}{5} ) $
C) $ \frac{1}{2}{{\tan }^{-1}}( \frac{3}{5} ) $
D) $ {{\tan }^{-1}}( \frac{1}{2} ) $
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Answer:
Correct Answer: A
Solution:
$ {{\tan }^{-1}}\frac{1}{4}+{{\tan }^{-1}}\frac{2}{9}={{\tan }^{-1}}( \frac{(1/4)+(2/9)}{1-(1/4)\times (2/9)} ) $ = $ {{\tan }^{-1}}( \frac{1}{2} )=\frac{1}{2}.2{{\tan }^{-1}}( \frac{1}{2} )=\frac{1}{2}{{\tan }^{-1}}\frac{2(1/2)}{1-(1/4)} $
$ =\frac{1}{2}{{\tan }^{-1}}\frac{4}{3}=\frac{1}{2}{{\cos }^{-1}}\frac{3}{5} $ .
BETA
