Question

Prove that: $\tan {70^ \circ } = \tan {20^ \circ } + 2\tan {50^ \circ }$

Answer 1

We will be needing certain properties of tan to solve this problem $\cot \theta = \tan \left( {{{90}^ \circ } - \theta } \right) = \dfrac{1}{{\tan \theta }}$ and also $\tan (\theta + \phi ) = \dfrac{{\tan \theta + \tan \phi }}{{1 - \tan \theta \tan \phi }}$ .

Complete step by step answer:
So we know that $\tan (\theta + \phi ) = \dfrac{{\tan \theta + \tan \phi }}{{1 - \tan \theta \tan \phi }}$ and also $\tan \left( {{{90}^ \circ } - \theta } \right) = \dfrac{1}{{\tan \theta }}$
So $\theta = {20^ \circ }\& \phi = {50^ \circ }$

\therefore \tan {70^ \circ } = \tan ({20^ \circ } + {50^ \circ }) = \dfrac{{\tan {{20}^ \circ } + \tan {{50}^ \circ }}}{{1 - \tan {{20}^ \circ }\tan {{50}^ \circ }}}\\\ \Rightarrow \tan {70^ \circ }(1 - \tan {20^ \circ }\tan {50^ \circ }) = \tan {20^ \circ } + \tan {50^ \circ }\\\ \Rightarrow \tan {70^ \circ } - \tan {70^ \circ }\tan {20^ \circ }\tan {50^ \circ } = \tan {20^ \circ } + \tan {50^ \circ }\\\ \end{array}$$ Now we can observe that $$\tan {70^ \circ }\tan {20^ \circ } = \tan ({90^ \circ } - {20^ \circ })\tan {20^ \circ } = 1$$ Putting this we get $$\begin{array}{l} \Rightarrow \tan {70^ \circ } - \tan {50^ \circ } = \tan {20^ \circ } + \tan {50^ \circ }\\\ \Rightarrow \tan {70^ \circ } = \tan {20^ \circ } + 2\tan {50^ \circ } \end{array}$$ Hence Proved. **Note:** Note that we can make a general formula, which would look something like; $$\tan \theta = \tan \left( {\dfrac{\pi }{2} - \theta } \right) + 2\tan \left( {2\theta - \dfrac{\pi }{2}} \right)$$ or equivalently $$\tan \theta = \tan \phi + 2\tan (\theta - \phi ),\forall \theta + \phi = \dfrac{\pi }{2},\theta > \phi $$.