Expressing $\tan 20°$ in terms of $\tan 35°$

Question

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?

Answer 1

If $\dfrac{1-a^2}{2a}=\dfrac{a-b}{1+ab}$

$$\iff\dfrac{a^3-3a}{1-3a^2}=\dfrac1b$$

If $a=\tan x,b=2-\sqrt3$

$$-\tan3x=\tan(-3x)=2+\sqrt3=\csc30^\circ+\cot30^\circ=\cdots=\cot15^\circ=\tan75^\circ$$

$\implies-3x=180^\circ n+75^\circ$ where $n$ is any integer

$\implies-x\equiv25^\circ,(60+25)^\circ,(120+25)^\circ\pmod{180^\circ}$

$\implies x\equiv-25^\circ\equiv155^\circ,-85^\circ\equiv95^\circ,-145^\circ\equiv35^\circ\pmod{180^\circ}$

Answer 2

$\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 $a$