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If $\sin^{-1}\frac{2a}{1+a^2} +\cos^{-1}\frac{1-a^2}{1+a^2}=\tan^{-1}\frac{2x}{1-x^2},where\; a,x,\in\;[0,1],$then the value of x is

$(a)\;0\qquad(b)\;\large\frac{a}{2}\qquad(c)\;a\qquad(d)\;\large\frac{2a}{1-a^{2}}$

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Toolbox:
  • Put \( a=tanA \Rightarrow A = tan^{-1}a\)
  • \( \large\frac{2tanA}{1+tan^2A}=sin2A\)
  • \( \large\frac{1-tan^2A}{1=tan^2A}=cos2A\)
  • \(2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2},\:\:\:|x|<1\)
Ans (D) \( \large\frac{2a}{1-a^2} \)
 
By taking a=tanA, \(tan^{-1}a=A\)
\(\large\frac{2a}{1+a^2}=\large\frac{2tanA}{1+tan^2A}=sin2A\:\:and\)
\(\large\frac{1-a^2}{1+a^2}= \large\frac{1-tan^2A}{1=tan^2A}=cos2A\)
\(sin^{-1}\large\frac{2a}{1+a^2}=sin^{-1}sin2A=2A\)
\(cos^{-1}\large\frac{1-a^2}{1+a^2}=cos^{-1}cos2A=2A\)
L.H.S. becomes
\(sin^{-1}\large\frac{2a}{1+a^2}+cos^{-1}\large\frac{1-a^2}{1+a^2}\)
\(=2A+2A=4A=4tan^{-1}a\)
Taking R.H.S., \(\tan^{-1}\large\frac{2x}{1-x^2}=2tan^{-1}x\)
Substituting the values of L.H.S. and R.H.S. in the given equation, we get
\(4tan^{-1}a=tan^{-1}\large \frac{2x}{1-x^2} =2tan^{-1}x\)
 
\( \Rightarrow 2tan^{-1}a=tan^{-1}x\)
By using the above formula of \( 2tan^{-1}x\)
 \( \Rightarrow\:2tan^{-1}a= tan^{-1} \large\frac{2a}{1-a^2}\)
\(\Rightarrow\:tan^{-1} \large\frac{2a}{1-a^2}=tan^{-1}x\)
\(\Rightarrow\: \large\frac{2a}{1-a^2}=x\)

 

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