Inverse Trigonometric Functions Question 232
Question: $ {{\sin }^{-1}}\frac{\sqrt{x}}{\sqrt{x+a}} $ is equal to
Options:
A) $ {{\cos }^{-1}}\sqrt{\frac{x}{a}} $
B) $ cose{c^{-1}}\sqrt{\frac{x}{a}} $
C) $ {{\tan }^{-1}}\sqrt{\frac{x}{a}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Putting $ x=a{{\tan }^{2}}\theta $
$ {{\sin }^{-1}}\frac{\sqrt{x}}{\sqrt{x+a}}={{\sin }^{-1}}\frac{\sqrt{a}\sqrt{{{\tan }^{2}}\theta }}{\sqrt{a{{\tan }^{2}}\theta +a}}={{\sin }^{-1}}\frac{\sqrt{a}\tan \theta }{\sqrt{a}\sec \theta } $
$ ={{\sin }^{-1}}\sin \theta =\theta ={{\tan }^{-1}}( \sqrt{\frac{x}{a}} ) $ .
BETA
