
Prove that: $ \tan 70^\circ = \tan 20^\circ + 2\tan 50^\circ $
Answer
563.4k+ views
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.
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.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

