A) \[{{\cos }^{-1}}\sqrt{\frac{x}{a}}\]
B) \[\text{cose}{{\text{c}}^{-1}}\sqrt{\frac{x}{a}}\]
C) \[{{\tan }^{-1}}\sqrt{\frac{x}{a}}\]
D) None of these
Correct Answer: C
Solution :
Putting \[x=a\,{{\tan }^{2}}\theta \] \[{{\sin }^{-1}}\frac{\sqrt{x}}{\sqrt{x+a}}={{\sin }^{-1}}\frac{\sqrt{a}\sqrt{{{\tan }^{2}}\theta }}{\sqrt{a\,{{\tan }^{2}}\theta +a}}={{\sin }^{-1}}\frac{\sqrt{a}\,\tan \theta }{\sqrt{a}\,\sec \theta }\] \[={{\sin }^{-1}}\sin \theta =\theta ={{\tan }^{-1}}\left( \sqrt{\frac{x}{a}} \right)\].
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