Inverse Trigonometric Functions Question 223
Question: $ {{\sin }^{-1}}\frac{4}{5}+2{{\tan }^{-1}}\frac{1}{3}= $
[ISM Dhanbad 1971]
Options:
A) $ \frac{\pi }{2} $
B) $ \frac{\pi }{3} $
C) $ \frac{\pi }{4} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{\sin }^{-1}}\frac{4}{5}={{\tan }^{-1}}\frac{4}{3},2{{\tan }^{-1}}\frac{1}{3}={{\tan }^{-1}}\frac{3}{4}={{\cot }^{-1}}\frac{4}{3} $ and $ {{\tan }^{-1}}x+{{\cot }^{-1}}x=\frac{\pi }{2} $ .
BETA
