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

 [/INST] $t=\arctan x\iff \tan t=x$.
Recall that $1+x^2=1+\tan^2 t=\frac{1}{\cos^2 t}$, so $\cos^2 t=\frac{1}{1+x^2}$.
Now, $$\cos(2\arctan x)=\cos 2t=\cos^2t-\sin^2t=2\cos^2t-1=2\frac{1}{1+x^2}-1=\frac{1-x^2}{1+x^2}$$ and it is a tautology.
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