Prove that$\frac{x+2-\sqrt{3}}{1+(\sqrt{3}-2)x}$=$\frac{1-{x^2}}{2x}$ Well, I was trying to find $\tan40$ in terms of $\tan25$=x.
So, I expanded 40 as 25+15 and got the value of $\tan40$ in terms of $x$ as $\frac{x+2-\sqrt{3}}{1+(\sqrt{3}-2)x}$.
Now, I solved $\tan40$ alternatively as 
$$\frac{\tan155-\tan115}{1+\tan155\tan115}= \frac{\tan(180-25)-tan(90+25)}{1+\tan(180-25)tan(90+25)}= \frac{-\tan25+\cot25}{1+\tan25{\cot25}}=\frac{1}{2}(-x+\frac{1}{x})=\frac{1-{x^2}}{2x}$$
But, we should have same solution from both the approaches. 
Therefore, $\frac{x+2-\sqrt{3}}{1+(\sqrt{3}-2)x}$ must be equals to $\frac{1-{x^2}}{2x}$. But I'm unable to prove it by algebra. So, can you kindly help me?

When $\frac{1-x^2}{2x}$ is equated to $\frac{x+2−\sqrt{3}}{1+(\sqrt{3}-2)x}$, it gives rise to: $$x^3-3(\sqrt{3}+2)x^2-3x+(\sqrt{3}+2)=0$$
Wolfram gives three roots:$$x≈-0.700208,\,\,\,x≈0.466308,\,\,\,x≈11.4301$$
Also, note that, $\tan{25}≈0.4663076...$
It is not a satisfying "exact" proof, but something to get a sense of things.
