The numerical value of \[{\left( {\tan 20^\circ + \tan 40^\circ + \sqrt 3 \tan 20^\circ \tan 40^\circ } \right)^2}\] is equal to:
Answer
627k+ views
Hint:
Here, we are required to find the numerical value of the given expression. We will use the formula of $\tan \left( {a + b} \right)$ to solve this. We will substitute the given measure of angle in the formula. Then by simplifying it further we will get the required value.
Formula Used:
We will use the formula $\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a \cdot \tan b}}$.
Complete step by step solution:
The given expression is \[{\left( {\tan 20^\circ + \tan 40^\circ + \sqrt 3 \tan 20^\circ \tan 40^\circ } \right)^2}\].
Now, as we know the formula of $\tan \left( {a + b} \right)$ is:
$\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a \cdot \tan b}}$…………………………………(1)
Substituting $a = 20^\circ $ and $b = 40^\circ $ in (1), we get
$\tan \left( {20^\circ + 40^\circ } \right) = \tan \left( {60^\circ } \right) = \dfrac{{\tan 20^\circ + \tan 40^\circ }}{{1 - \tan 20^\circ .\tan 40^\circ }}$…………………..(2)
Now, we know that $\tan 60^\circ = \sqrt 3 $.
Hence, substituting the value $\tan 60^\circ = \sqrt 3 $ in (2), we get
$ \Rightarrow \sqrt 3 = \dfrac{{\tan 20^\circ + \tan 40^\circ }}{{1 - \tan 20^\circ \cdot \tan 40^\circ }}$
On cross-multiplication, we get
$ \Rightarrow \sqrt 3 - \sqrt 3 \tan 20^\circ \cdot \tan 40^\circ = \tan 20^\circ + \tan 40^\circ $
Adding $\sqrt 3 \tan 20^\circ \cdot \tan 40^\circ $ on both sides, we get,
$ \Rightarrow \sqrt 3 = \tan 20^\circ + \tan 40^\circ + \sqrt 3 \tan 20^\circ \cdot \tan 40^\circ $
Now, squaring both sides, we get
$ \Rightarrow {\left( {\tan 20^\circ + \tan 40^\circ + \sqrt 3 \tan 20^\circ \cdot \tan 40^\circ } \right)^2} = {\left( {\sqrt 3 } \right)^2} = 3$
Therefore, the numerical value of \[{\left( {\tan 20^\circ + \tan 40^\circ + \sqrt 3 \tan 20^\circ \tan 40^\circ } \right)^2}\] must be 3.
Note:
Trigonometry is a branch of mathematics that helps us to study the relationship between the sides and the angles of a triangle. In practical life, trigonometry is used by cartographers (to make maps). It is also used by the aviation and naval industries. In fact, trigonometry is even used by Astronomers to find the distance between two stars. Hence, it has an important role to play in everyday life. The three most common trigonometric functions are the tangent function, the sine and the cosine function. In simple terms they are written as ‘sin’, ‘cos’, and ‘tan’. Hence, trigonometry is not just a chapter to study, in fact, it is being used in everyday life.
Here, we are required to find the numerical value of the given expression. We will use the formula of $\tan \left( {a + b} \right)$ to solve this. We will substitute the given measure of angle in the formula. Then by simplifying it further we will get the required value.
Formula Used:
We will use the formula $\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a \cdot \tan b}}$.
Complete step by step solution:
The given expression is \[{\left( {\tan 20^\circ + \tan 40^\circ + \sqrt 3 \tan 20^\circ \tan 40^\circ } \right)^2}\].
Now, as we know the formula of $\tan \left( {a + b} \right)$ is:
$\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a \cdot \tan b}}$…………………………………(1)
Substituting $a = 20^\circ $ and $b = 40^\circ $ in (1), we get
$\tan \left( {20^\circ + 40^\circ } \right) = \tan \left( {60^\circ } \right) = \dfrac{{\tan 20^\circ + \tan 40^\circ }}{{1 - \tan 20^\circ .\tan 40^\circ }}$…………………..(2)
Now, we know that $\tan 60^\circ = \sqrt 3 $.
Hence, substituting the value $\tan 60^\circ = \sqrt 3 $ in (2), we get
$ \Rightarrow \sqrt 3 = \dfrac{{\tan 20^\circ + \tan 40^\circ }}{{1 - \tan 20^\circ \cdot \tan 40^\circ }}$
On cross-multiplication, we get
$ \Rightarrow \sqrt 3 - \sqrt 3 \tan 20^\circ \cdot \tan 40^\circ = \tan 20^\circ + \tan 40^\circ $
Adding $\sqrt 3 \tan 20^\circ \cdot \tan 40^\circ $ on both sides, we get,
$ \Rightarrow \sqrt 3 = \tan 20^\circ + \tan 40^\circ + \sqrt 3 \tan 20^\circ \cdot \tan 40^\circ $
Now, squaring both sides, we get
$ \Rightarrow {\left( {\tan 20^\circ + \tan 40^\circ + \sqrt 3 \tan 20^\circ \cdot \tan 40^\circ } \right)^2} = {\left( {\sqrt 3 } \right)^2} = 3$
Therefore, the numerical value of \[{\left( {\tan 20^\circ + \tan 40^\circ + \sqrt 3 \tan 20^\circ \tan 40^\circ } \right)^2}\] must be 3.
Note:
Trigonometry is a branch of mathematics that helps us to study the relationship between the sides and the angles of a triangle. In practical life, trigonometry is used by cartographers (to make maps). It is also used by the aviation and naval industries. In fact, trigonometry is even used by Astronomers to find the distance between two stars. Hence, it has an important role to play in everyday life. The three most common trigonometric functions are the tangent function, the sine and the cosine function. In simple terms they are written as ‘sin’, ‘cos’, and ‘tan’. Hence, trigonometry is not just a chapter to study, in fact, it is being used in everyday life.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

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

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths 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

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

The teeth used for biting and cutting food are called class 11 biology CBSE

