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