<s>[INST] 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.
 [/INST] You can use the identity $2 \cot 2 A = \cot A - \tan A$ which makes the job easy.
$\tan 20° + 2\tan 50° = \tan 20° + 2\cot 40°$
$= \tan 20° + \cot 20° - \tan 20^0 = \cot 20^0 = \tan 70^0$
Otherwise proceed from $\tan 20° + 2\tan 50° = \tan 20° + 2\cot 40°$ as below -
$\displaystyle \tan 20^0 + 2 \cot 40^0 = \tan 20° + 2 \cdot \frac{\cos^2 20^0 - \sin^2 20^0}{2 \cos 20^0 \sin 20^0}$
$= \tan 20^0 + \cot 20^0 - \tan 20^0 = \tan 70^0$
</s>