Showing $\tan70° = \tan20° + 2\tan50°$ Q. Prove that $\tan 70° = \tan 20° + 2\tan 50°$.
My approach:
$ LHS = \tan 70° = \dfrac1{\cot 70°} = \dfrac1{\tan 20°}$
$ \begin{align}
RHS 
&= \tan 20° + 2\tan 50° \\
&=  \tan 20° + 2\tan (20+30)° \\
&= \tan 20° + \dfrac{2(\tan 20° + 1/√3)}{1 - \tan 20°/√3} \\
&=\dfrac{2 + 3√3 \tan 20° - \tan^2 20°)}{√3 - \tan 20°} 
\end{align}
$
Why are the two sides not equal despite being expressed in the same terms? Can someone offer some help?
Much to my surprise, my friend just expanded tan70° and cross-multiplied the terms to prove it.

This will answer the question you asked in bold.
In fact, the two expressions you got involving $\tan 20^\circ$ are equal, not in the sense that algebraic manipulation will produce identical expressions, but rather depending upon a special fact regarding $\tan 20^\circ$. You want to show for $x=\tan 20^\circ$,  $$\frac 1x = \frac{2+3\sqrt 3 x -x^2}{\sqrt 3-x}\\ \iff x^3-3\sqrt 3x^2 -3x +\sqrt 3 =0 $$
To do this, invoke the fact that $3\times 20 =60$ and the formula for $\tan 3x$: $$\sqrt 3 = \frac{3x -x^3}{1-3x^2} $$ and this rearranges to the same cubic equation above.
