Inverse Trigonometric Functions Question 206
Question: $ \cos [ {{\tan }^{-1}}\frac{1}{3}+{{\tan }^{-1}}\frac{1}{2} ]= $
[MP PET 1991; MNR 1990]
Options:
A) $ \frac{1}{\sqrt{2}} $
B) $ \frac{\sqrt{3}}{2} $
C) $ \frac{1}{2} $
D) $ \frac{\pi }{4} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \cos [ {{\tan }^{-1}}\frac{1}{3}+{{\tan }^{-1}}\frac{1}{2} ]=\cos [ {{\tan }^{-1}}( \frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3}\times \frac{1}{2}} ) ] $
$ =\cos {{{\tan }^{-1}}(1)}=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}} $ .
BETA
