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
636k+ 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
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

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

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

What do you mean by retardation What is its SI uni 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

Difference between physical and chemical change class 11 chemistry CBSE

