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Choose the correct answer \(sin (tan^{-1}x), |x| <1 \)is equal to

$(A)\;\frac{x}{\sqrt{1-x^2}} \\(B)\;\frac{1}{\sqrt{1-x^2}} \\(C)\;\frac{1}{\sqrt{1+x^2}} \\ (D)\;\frac{x}{\sqrt{1+x^2}}$
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Toolbox:
  • If \(\tan \theta = x\) then \(\sin \theta = \large \frac {x}{\sqrt(1+x^2)} \)
Ans: (D) \( \frac{x}{\sqrt{1+x^2}}\)
Let \( tan^{-1}x=\theta\) \( \Rightarrow tan\theta = x \Rightarrow sin\theta = \large\frac{x}{\sqrt{1+x^2}}\)
Substituting for $\theta$, we get, \( sin(tan^{-1}x)=sin\theta=\large \frac{x}{\sqrt{1+x^2}}\)
 

 

answered Feb 23, 2013 by thanvigandhi_1
edited Mar 15, 2013 by rvidyagovindarajan_1
 

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