Inverse Trigonometric Functions Question 46
Question: $ \tan ({{\cos }^{-1}}x) $ is equal to
[IIT 1993]
Options:
A) $ \frac{\sqrt{1-x^{2}}}{x} $
B) $ \frac{x}{1+x^{2}} $
C) $ \frac{\sqrt{1+x^{2}}}{x} $
D) $ \sqrt{1-x^{2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ {{\cos }^{-1}}x=\theta . $ Then $ x=\cos \theta $
$ \Rightarrow \tan \theta =\sqrt{{{\sec }^{2}}\theta -1} $
$ =\sqrt{\frac{1}{x^{2}}-1}=\sqrt{\frac{1-x^{2}}{x}} $
$ \therefore \tan ({{\cos }^{-1}}x)=\tan \theta =\frac{\sqrt{1-x^{2}}}{x} $ .
BETA
