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

Toolbox:

• $cosy=\large\frac{1}{\sqrt{1+tan^2y}}$
• $sin2A=2sinAcosA$
• $tan^{-1}\theta=cot^{-1}\large\frac{1}{\theta}$
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
$sin2\alpha = 2sin\alpha.cos\alpha$
$=2.\large\frac{\sqrt{1+x^2}}{\sqrt{2+x^2}}.\large\frac{1}{\sqrt{2+x^2}}$
$= \large\frac{2\sqrt{1+x^2}}{2+x^2}$

edited Mar 19, 2013