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?

On rearrange we have
$$2+\sqrt3=\dfrac{3x-x^3}{1-3x^2}=\cdots=\tan(3\cdot25^\circ)$$
Now $$\tan(75^\circ)=\tan(45^\circ+30^\circ)=?$$
