Expressing $\tan 20°$ in terms of $\tan 35°$ If $\tan 35^\circ = a$, we are required to express $\left(\frac{\tan 145^\circ - \tan 125^\circ}{1 + \tan 145^\circ\tan 125^\circ}\right)$ in terms of $a$.
Here's one way to solve this:  $$\frac{\tan 145^\circ - \tan 125^\circ}{1 + \tan 145^\circ\tan 125^\circ} = \tan (145^\circ - 125^\circ) = \tan 20^\circ = \tan (90^\circ - 70^\circ) = \cot 70^\circ = \frac{1}{\tan 70^\circ} = \frac{1}{\tan (2 \times 35^\circ)} = \frac{1 - \tan^2 35^\circ}{2\tan 35^\circ} = \frac{1 - a^2}{2a}$$
However, I tried to solve it using another method as described below, and faced a problem:
$$\frac{\tan 145^\circ - \tan 125^\circ}{1 + \tan 145^\circ\tan 125^\circ} = \tan 20^\circ = \tan (35^\circ - 15^\circ) = \frac{\tan 35^\circ - \tan15^\circ}{1 + \tan 35^\circ\tan 15^\circ} = \frac{a - (2 - \sqrt3)}{1 + a(2 - \sqrt3)} = \frac{a - 2 + \sqrt3}{1 + 2a - \sqrt3a}$$
I tried to simplify it to get $\frac{1 - a^2}{2a}$, but I couldn't. So my question is, is there any way to show that $\frac{a - 2 + \sqrt3}{1 + 2a - \sqrt3a}$ is equal to $\frac{1 - a^2}{2a}$? If not, why are we getting two different answers?

$\tan145^\circ=\tan(180^\circ-35^\circ) =-\tan35^\circ $ and $\tan125^\circ=\tan(90^\circ+35^\circ) =-\frac{1}{\tan35^\circ} $
For your two answers, have you find their values with a calculator?
Actually, you have proven an equality in $a$
