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If $|\;x\;|\leq 1,then\;2\tan^{-1}x+\sin^{-1}\frac{2x}{1+x^2}$ is equal to:

\[(A)\quad4\tan^{-1}x\quad(B)\quad 0\quad(C)\quad\frac{1}{2}\quad(D)\quad \pi\]
Can you answer this question?
 
 

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  • \( \large\frac{2tan\theta}{1+tan^2\theta}=sin2\theta\)
Ans - (A) \( 4tan^{-1}x\)
 
put \( x=tan\theta \Rightarrow \theta=tan^{-1}x\)
\(\large\frac{2x}{1+x^2}=\large\frac{2tan\theta}{1+tan^2\theta}=sin2\theta\)
\(\Rightarrow\:sin^{-1}\large\frac{2x}{1+x^2}=sin^{-1}sin2\theta=2\theta\)
 
Substituting the value of x, the given expression becomes
\(2tan^{-1}tan\theta+sin^{-1}\large\frac{2tan\theta}{1+tan^\theta}\)
 
\( \Rightarrow 2\theta+2\theta=4\theta\)
\(=4tan^{-1}x\)

 

answered Feb 18, 2013 by thanvigandhi_1
edited Mar 16, 2013 by thanvigandhi_1
 

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