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Prove that $\sin2[\cot^{-1}\{\cos(\tan^{-1}x)\}]=2\frac{\sqrt{x^2+1}}{x^2+2}$

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  • \( cosy=\large\frac{1}{\sqrt{1+tan^2y}}\)
  • \( sin2A=2sinAcosA\)
  • \(tan^{-1}\theta=cot^{-1}\large\frac{1}{\theta}\)
put \( tan^{-1}x=y,\:\:\Rightarrow\: x=tany\)
\( sin2 \bigg[ cot^{-1} (cosy) \bigg] \)
 
From the above formula of cosy
\( cosy=\large\frac{1}{\sqrt{1+tan^2y}}=\large\frac{1}{\sqrt{1+x^2}}\)
\( \Rightarrow sin2 \bigg( cot^{-1} \bigg(\large \frac{1}{\sqrt{1+x^2}} \bigg) \bigg) \)
 
By taking \(\theta=\sqrt{1+x^2}\) in the above formula
\(cot^{-1}\large\frac{1}{\sqrt{1+x^2}}=tan^{-1}\sqrt{1+x^2}\)
\( \Rightarrow sin2 \: tan^{-1} \bigg( \sqrt{1+x^2} \bigg) \)
Put \( tan^{-1} ( \sqrt{1+x^2} ) = \alpha \)
\(\Rightarrow\:tan\alpha=\sqrt{1+x^2}\)
 
We know that in a right angled triangle \(tan\alpha=\large\frac{perpendicular}{base}=\large\frac{\sqrt{1+x^2}}{1}\)
and \((base)^2+(perpendicular)^2=(hypotenuse)^2\) from pythogorous theorem
\(\Rightarrow\:1+(1+x^2)=2+x^2\)
\(\Rightarrow\:sin\alpha=\large\frac{perpendicular}{hypotenuse}=\large\frac{\sqrt{1+x^2}}{\sqrt{2+x^2}}\), and
\(cos\alpha=\large\frac{base}{hypotenuse}=\large\frac{1}{\sqrt{2+x^2}}\)
 
Substituting the values, the given expression becomes
\( sin2\alpha = 2sin\alpha.cos\alpha\)
\(=2.\large\frac{\sqrt{1+x^2}}{\sqrt{2+x^2}}.\large\frac{1}{\sqrt{2+x^2}}\)
\( = \large\frac{2\sqrt{1+x^2}}{2+x^2} \)

 

answered Feb 19, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 
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