Inverse Trigonometric Functions Question 200
Question: $ \cos
[ 2{{\cos }^{-1}}\frac{1}{5}+{{\sin }^{-1}}\frac{1}{5} ]= $ [IIT 1981]
Options:
A) $ \frac{2\sqrt{6}}{5} $
B) $ -\frac{2\sqrt{6}}{5} $
C) $ \frac{1}{5} $
D) $ -\frac{1}{5} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \cos ( {{\cos }^{-1}}\frac{1}{5}+{{\sin }^{-1}}\frac{1}{5}+{{\cos }^{-1}}\frac{1}{5} )=\cos ( \frac{\pi }{2}+{{\cos }^{-1}}\frac{1}{5} ) $
$ =-\sin ( {{\cos }^{-1}}\frac{1}{5} )=-\sin ( {{\sin }^{-1}}\sqrt{\frac{24}{25}} )=-\frac{2\sqrt{6}}{5} $ .
BETA
