Inverse Trigonometric Functions Question 179
Question: $ {{\sin }^{-1}}( \frac{3}{5} )+{{\tan }^{-1}}( \frac{1}{7} )= $
[Karnataka CET 1994]
Options:
A) $ \frac{\pi }{4} $
B) $ \frac{\pi }{2} $
C) $ {{\cos }^{-1}}( \frac{4}{5} ) $
D) $ \pi $
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Answer:
Correct Answer: A
Solution:
$ {{\sin }^{-1}}\frac{3}{5}+{{\tan }^{-1}}\frac{1}{7}={{\tan }^{-1}}\frac{3}{4}+{{\tan }^{-1}}\frac{1}{7} $
$ ={{\tan }^{-1}}( \frac{(3/4)+(1/7)}{1-(3/4)\times (1/7)} )={{\tan }^{-1}}( \frac{25}{25} )={{\tan }^{-1}}1=\frac{\pi }{4} $ .
BETA
