A) \[\frac{1}{\sqrt{1+{{x}^{2}}}}\]
B) \[\frac{x}{\sqrt{1+{{x}^{2}}}}\]
C) \[-\frac{1}{\sqrt{1+{{x}^{2}}}}\]
D) \[-\frac{x}{\sqrt{1+{{x}^{2}}}}\]
Correct Answer: B
Solution :
Let \[{{\tan }^{-1}}x=\theta \] \[\Rightarrow \] \[\tan \theta =x\] \[\because \]\[\sin \,({{\tan }^{-1}}x)=\sin \theta \] \[=\frac{BC}{AC}=\frac{x}{\sqrt{1+{{x}^{2}}}}\]You need to login to perform this action.
You will be redirected in
3 sec