Inverse Trigonometric Functions Question 154
Question: $ \tan [ {{\sec }^{-1}}\sqrt{1+x^{2}} ]= $
Options:
A) $ \frac{1}{x} $
B) x
C) $ \frac{1}{\sqrt{1+x^{2}}} $
D) $ \frac{x}{\sqrt{1+x^{2}}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \tan ( {{\sec }^{-1}}\sqrt{1+x^{2}} )=\tan ( {{\sec }^{-1}}\sqrt{1+{{\tan }^{2}}\theta } ) $ (Putting $ x=\tan \theta ) $
$ =\tan ({{\sec }^{-1}}\sec \theta )=\tan \theta =x $ .
BETA
