question_answer

\[{{\tan }^{-1}}\,\frac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}}\]is equal to

A)  \[2{{\sin }^{-1}}\frac{x}{a}\]

B)  \[{{\sin }^{-1}}\frac{2x}{a}\]

C)  \[{{\sin }^{-1}}\frac{x}{a}\]

D)  \[{{\cos }^{-1}}\frac{x}{a}\]

Correct Answer: C

Solution :

Let  \[{{\tan }^{-1}}\,\frac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}}=\theta \] \[\Rightarrow \] \[\tan \theta =\frac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}}\] \[\therefore \] \[A{{C}^{2}}={{x}^{2}}+{{a}^{2}}-{{x}^{2}}\] \[\Rightarrow \] \[AC=a\] \[\therefore \] \[\sin \theta =\frac{x}{a}\]