Question

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

Answer 1

Hint: 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 \[\theta = {20^ \circ }\& \phi = {50^ \circ }\]
\[\begin{array}{l}
\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 \[\tan \theta = \tan \phi + 2\tan (\theta - \phi ),\forall \theta + \phi = \dfrac{\pi }{2},\theta > \phi \].