Prove that ${\tan70^0} - \tan {20^0} = 2\tan {50^0}$

Hint: We can rewrite $\tan \left( {{{70}^0}} \right)$ as $\tan \left( {{{50}^0} + {{20}^0}} \right)$, right? For the same, let’s apply tan(a+b) formula and equate it to $\tan \left( {{{70}^0}} \right)$. Then, with simple simplification we’ll get the answer.

Complete answer:

We know that tan (a + b) = $\dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}}$

$\tan \left( {{{70}^0}} \right) = \tan \left( {{{50}^0} + {{20}^0}} \right)$

So using above formulae

${\text{tan7}}{{\text{0}}^0} = \dfrac{{\tan {{50}^0} + \tan {{20}^0}}}{{1 - \tan {{50}^0}\tan {{20}^0}}}$

Or ${\text{ if we cross multiply we get}}$

${\text{tan7}}{0^0} - \tan {70^0}\tan {50^0}\tan {20^0} = \tan {50^0} + \tan {20^0}$

Or ${\text{ tan7}}{0^0} - \tan {20^0} = \tan {50^0} + \tan {70^0}\tan {50^0}\tan {20^0}.........\left\{ 1 \right\}$

Now, let's solve for $\tan {70^0}\tan {50^0}\tan {20^0}$

We can write it as $\tan {50^0}\tan ({90^0} - {20^0})\tan {20^0}$

Which is equal to $\tan {50^0}\cot {20^0}\tan {20^0} = \tan {50^0}$

Hence equation 1 gets changed to

${\text{tan7}}{0^0} - \tan {20^0} = \tan {50^0} + \tan {50^0} = 2\tan {50^0}$

Hence proved.

Note - Always start such types of proofs by thinking how we can break the LHS part in terms of angles in RHS then proceed further with the respective formulas.