Inverse Trigonometric Functions Question 153
Question: During $ \cos ({{\tan }^{-1}}x)= $
[MP PET 1988; MNR 1981]
Options:
A) $ \sqrt{1+x^{2}} $
B) $ \frac{1}{\sqrt{1+x^{2}}} $
C) $ 1+x^{2} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ \theta ={{\tan }^{-1}}x\Rightarrow x=\tan \theta $
$ \therefore \cos \theta =\frac{1}{\sqrt{1+{{\tan }^{2}}\theta }}=\frac{1}{\sqrt{1+x^{2}}} $ Hence $ \cos \theta =\cos ({{\tan }^{-1}}x)=\frac{1}{\sqrt{1+x^{2}}} $ .
BETA
