Question

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

Answer 1

Hint: Write $ {70^0} $ as the sum of $ {20^0} $ and $ {50^0}. $ Then use the expansion formula of $ \tan $ to solve this question. One must be aware how to use allied angles to arrive at the result.

Complete step-by-step answer:
We will use the formula,
 $ \tan (\theta + \phi ) = \dfrac{{\tan \theta + \tan \phi }}{{1 - \tan \theta \tan \phi }} $ . . . (1)
We can write
 $ \tan 70^\circ = \tan (20^\circ + 50^\circ ) $
So, by using equation (1), we can write
 $ \tan 70^\circ = \tan (20^\circ + 50^\circ ) $
 $ \Rightarrow \tan 70^\circ = \dfrac{{\tan 20^\circ + \tan 50^\circ }}{{1 - \tan 20^\circ \tan 50^\circ }} $
By rearranging, we get
\[\tan 70^\circ \times (1 - \tan 20^\circ \tan 50^\circ ) = \tan 20^\circ + \tan 50^\circ \]
Expanding the bracket, we get
\[\tan 70^\circ - \tan 20^\circ \tan 50^\circ \tan 70^\circ = \tan 20^\circ + \tan 50^\circ \]
 $ \Rightarrow \tan 70^\circ - \tan 20^\circ \tan 50^\circ \tan (90^\circ - {20^0}) = \tan 20^\circ + \tan 50^\circ $
\[ \Rightarrow \tan 70^\circ - \tan 20^\circ \tan 50^\circ \cot 20^\circ = \tan 20^\circ + \tan 50^\circ \] $ \left( {\because \tan ({{90}^0} - \theta ) = \cot \theta } \right) $
\[ \Rightarrow \tan 70^\circ - \tan 20^\circ \tan 50^\circ \dfrac{1}{{\tan 20^\circ }} = \tan 20^\circ + \tan 50^\circ \] $ \left( {\because \cot \theta = \dfrac{1}{{\tan \theta }}} \right) $
Cancel the common terms
\[ \Rightarrow \tan 70^\circ - \tan 50^\circ = \tan 20^\circ + \tan 50^\circ \]
\[ \Rightarrow \tan 70^\circ = \tan 20^\circ + 2\tan 50^\circ \]
Hence proved.

Note: Observe the question carefully. You should understand that the number is LHS is the sum of numbers in RHS. That is when it should occur to you that you have to write $ {70^0} = {50^0} + {20^0} $
Finding a relation in the terms of question is important in such types of questions. Otherwise, it would be difficult to understand which formula to use to solve the question.