A) \[\sqrt{1+{{x}^{2}}}\]
B) \[\frac{1}{\sqrt{1+{{x}^{2}}}}\]
C) \[1+{{x}^{2}}\]
D) None of these
Correct Answer: B
Solution :
Let \[\theta ={{\tan }^{-1}}x\,\,\Rightarrow \,\,x=\tan \theta \] \[\therefore \,\,\cos \theta =\frac{1}{\sqrt{1+{{\tan }^{2}}\theta }}=\frac{1}{\sqrt{1+{{x}^{2}}}}\] Hence\[\cos \theta =\cos \,({{\tan }^{-1}}x)=\frac{1}{\sqrt{1+{{x}^{2}}}}\].
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