Prove that$\frac{x+2-\sqrt{3}}{1+(\sqrt{3}-2)x}$=$\frac{1-{x^2}}{2x}$

Question

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?

Answer 1

On rearrange we have

$$2+\sqrt3=\dfrac{3x-x^3}{1-3x^2}=\cdots=\tan(3\cdot25^\circ)$$

Now $$\tan(75^\circ)=\tan(45^\circ+30^\circ)=?$$

Answer 2

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.