Inverse Trigonometric Functions Question 25
Question: $ \sin ( 4{{\tan }^{-1}}\frac{1}{3} )= $
Options:
A) $ \frac{12}{25} $
B) $ \frac{24}{25} $
C) $ \frac{1}{5} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \sin ( 4{{\tan }^{-1}}\frac{1}{3} )=\sin [ 2{{\tan }^{-1}}( \frac{2/3}{1-(1/9)} ) ] $
$ =\sin [ 2{{\tan }^{-1}}\frac{3}{4} ]=\sin {{\sin }^{-1}}( \frac{2\times (3/4)}{1+(9/16)} ) $
$ =\frac{3}{2}\times \frac{16}{25}=\frac{24}{25} $
$ ( \because 2{{\tan }^{-1}}x={{\sin }^{-1}}\frac{2x}{1+x^{2}} ) $
BETA
