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 \[{{\tan }^{-1}}\,\frac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}}=\theta \] \[\Rightarrow \] \[\tan \theta =\frac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}}\] \[\therefore \] \[A{{C}^{2}}={{x}^{2}}+{{a}^{2}}-{{x}^{2}}\] \[\Rightarrow \] \[AC=a\] \[\therefore \] \[\sin \theta =\frac{x}{a}\]You need to login to perform this action.
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