General solution of $\cos (2\arctan x) = \frac{{1 - x^2 }}{{1 + x^2 }}?$ 
How to find the general solution of $$\cos (2\arctan x) = \frac{{1 - x^2 }}{{1 + x^2 }}$$


You can start by substituting both sides into $\arccos()$ function:
$$\arccos(\cos(2 \arctan x)) = \arccos\Biggl(\frac{1-x^2}{1+x^2}\Biggl)$$ $$2 \arctan x =\arccos\Biggl(\frac{1-x^2}{1+x^2}\Biggl)$$
Then you can take the derivative of both sides: $$\begin{align} \ \frac{2}{1+x^2} &=\ \frac{1}{\sqrt{1-\Biggl(\frac{1-x^2}{1+x^2}\Biggl)^2}} \frac{4x}{(1+x^2)^2} \\
&=\ \sqrt{\frac{(1+x^2)^2}{4x^2}} \frac{4x}{(1+x^2)^2} \\ \end{align}$$
If you organize the right-hand side of the equation, you eventually get:
$$\bbox[yellow] {\frac{2}{1+x^2}=\frac{2}{1+x^2}}$$
This means the equation holds for all real $x$ values.
