Prove the following trigonometric equation:
\[\tan {{20}^{o}}.\tan {{40}^{o}}.\tan {{60}^{o}}.\tan {{80}^{o}}=3\]
Last updated date: 17th Mar 2023
•
Total views: 306k
•
Views today: 8.88k
Answer
306k+ views
Hint: First of all, as we know that we can substitute \[{{40}^{o}}=\left( {{60}^{o}}-{{20}^{o}} \right)\] and \[{{80}^{o}}=\left( {{60}^{o}}+{{20}^{o}} \right)\]. Then use the formula of \[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}\] and \[\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}\].
The expression in the question to be proved is given as
\[\tan {{20}^{o}}.\tan {{40}^{o}}.\tan {{60}^{o}}.\tan {{80}^{o}}=3\]
Let us consider the LHS of the given expression as below,
\[A=\tan {{20}^{o}}.\tan {{40}^{o}}.\tan {{60}^{o}}.\tan {{80}^{o}}\]
Since, we know that,
\[\tan {{60}^{o}}=\sqrt{3}\]
Therefore, by putting the value of \[\tan {{60}^{o}}\] in the above expression, we get,
\[A=\tan {{20}^{o}}.\tan {{40}^{o}}.\sqrt{3}.\tan {{80}^{o}}\]
We can also write the above expression as
\[A=\sqrt{3}.\tan {{20}^{o}}.\tan {{40}^{o}}.\tan {{80}^{o}}\]
Now, we can know that \[{{40}^{o}}=\left( {{60}^{o}}-{{20}^{o}} \right)\] and \[{{80}^{o}}=\left( {{60}^{o}}+{{20}^{o}} \right)\].
So, we get the above expression as,
\[A=\sqrt{3}.\tan {{20}^{o}}.\tan \left( {{60}^{o}}-{{20}^{o}} \right).\tan \left( {{60}^{o}}+{{20}^{o}} \right)\]
Since, we know that
\[\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A.\tan B}\]
And,
\[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A.\tan B}\]
Therefore, by applying the above formulas, we get the above expression as
\[A=\left( \sqrt{3} \right).\left( \tan {{20}^{o}} \right)\left[ \dfrac{\tan {{60}^{o}}-\tan {{20}^{o}}}{1+\tan {{60}^{o}}.\tan {{20}^{o}}} \right].\left[ \dfrac{\tan {{60}^{o}}+\tan {{20}^{o}}}{1-\tan {{60}^{o}}\tan {{20}^{o}}} \right]\]
Since, we know that,
\[\tan {{60}^{o}}=\sqrt{3}\]
Therefore, by putting the value of \[\tan {{60}^{o}}\] in the above expression, we get
\[A=\left( \sqrt{3} \right).\left( \tan {{20}^{o}} \right)\left[ \dfrac{\sqrt{3}-\tan {{20}^{o}}}{1+\sqrt{3}.\tan {{20}^{o}}} \right].\left[ \dfrac{\sqrt{3}+\tan {{20}^{o}}}{1-\sqrt{3}\tan {{20}^{o}}} \right]\]
We can also write the above expression as,
\[A=\dfrac{\sqrt{3}\tan {{20}^{o}}.\left( \sqrt{3}-\tan {{20}^{o}} \right)\left( \sqrt{3}+\tan {{20}^{o}} \right)}{\left( 1-\sqrt{3}.\tan {{20}^{o}} \right)\left( 1+\sqrt{3}\tan {{20}^{o}} \right)}\]
Since we know that
\[\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}\]
Therefore, by applying this formula in the above expression, we get,
\[A=\sqrt{3}\left( \tan {{20}^{o}} \right).\left[ \dfrac{{{\left( \sqrt{3} \right)}^{2}}-{{\left( \tan {{20}^{o}} \right)}^{2}}}{{{\left( 1 \right)}^{2}}-{{\left( \sqrt{3}\tan {{20}^{o}} \right)}^{2}}} \right]\]
By simplifying the above equation, we get,
\[\Rightarrow A=\sqrt{3}\left( \tan {{20}^{o}} \right).\left[ \dfrac{3-{{\tan }^{2}}{{20}^{o}}}{1-3{{\tan }^{2}}{{20}^{o}}} \right]\]
Now, we will take \[\tan {{20}^{o}}\] inside the bracket
We get,
\[\Rightarrow A=\sqrt{3}\left[ \dfrac{3\tan {{20}^{o}}-{{\tan }^{3}}{{20}^{o}}}{1-3{{\tan }^{2}}{{20}^{o}}} \right]\]
Since, we know that
\[\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta }=\tan 3\theta \]
Therefore, by applying the above formula, we get,
\[A=\sqrt{3}\left[ \tan 3.\left( {{20}^{o}} \right) \right]\]
We can also write the above expression as
\[A=\sqrt{3}\left[ \tan {{60}^{o}} \right]\]
Since, we know that \[\tan {{60}^{o}}=\sqrt{3}\], therefore by putting the value of \[\tan {{60}^{o}}\] in the above expression we get,
\[A=\sqrt{3}.\sqrt{3}\]
Therefore, A = 3 = RHS
Hence, we proved that the value of \[\tan {{20}^{o}}\tan {{40}^{o}}\tan {{60}^{o}}\tan {{80}^{o}}=3\].
Note: Here by looking at the terms like \[\tan {{20}^{o}},\tan {{40}^{o}}\] and \[\tan {{60}^{o}}\], students often make this mistake of using formulas of double angles that is \[\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }\] which makes the solution lengthy and does not lead to the desired result.
The expression in the question to be proved is given as
\[\tan {{20}^{o}}.\tan {{40}^{o}}.\tan {{60}^{o}}.\tan {{80}^{o}}=3\]
Let us consider the LHS of the given expression as below,
\[A=\tan {{20}^{o}}.\tan {{40}^{o}}.\tan {{60}^{o}}.\tan {{80}^{o}}\]
Since, we know that,
\[\tan {{60}^{o}}=\sqrt{3}\]
Therefore, by putting the value of \[\tan {{60}^{o}}\] in the above expression, we get,
\[A=\tan {{20}^{o}}.\tan {{40}^{o}}.\sqrt{3}.\tan {{80}^{o}}\]
We can also write the above expression as
\[A=\sqrt{3}.\tan {{20}^{o}}.\tan {{40}^{o}}.\tan {{80}^{o}}\]
Now, we can know that \[{{40}^{o}}=\left( {{60}^{o}}-{{20}^{o}} \right)\] and \[{{80}^{o}}=\left( {{60}^{o}}+{{20}^{o}} \right)\].
So, we get the above expression as,
\[A=\sqrt{3}.\tan {{20}^{o}}.\tan \left( {{60}^{o}}-{{20}^{o}} \right).\tan \left( {{60}^{o}}+{{20}^{o}} \right)\]
Since, we know that
\[\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A.\tan B}\]
And,
\[\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A.\tan B}\]
Therefore, by applying the above formulas, we get the above expression as
\[A=\left( \sqrt{3} \right).\left( \tan {{20}^{o}} \right)\left[ \dfrac{\tan {{60}^{o}}-\tan {{20}^{o}}}{1+\tan {{60}^{o}}.\tan {{20}^{o}}} \right].\left[ \dfrac{\tan {{60}^{o}}+\tan {{20}^{o}}}{1-\tan {{60}^{o}}\tan {{20}^{o}}} \right]\]
Since, we know that,
\[\tan {{60}^{o}}=\sqrt{3}\]
Therefore, by putting the value of \[\tan {{60}^{o}}\] in the above expression, we get
\[A=\left( \sqrt{3} \right).\left( \tan {{20}^{o}} \right)\left[ \dfrac{\sqrt{3}-\tan {{20}^{o}}}{1+\sqrt{3}.\tan {{20}^{o}}} \right].\left[ \dfrac{\sqrt{3}+\tan {{20}^{o}}}{1-\sqrt{3}\tan {{20}^{o}}} \right]\]
We can also write the above expression as,
\[A=\dfrac{\sqrt{3}\tan {{20}^{o}}.\left( \sqrt{3}-\tan {{20}^{o}} \right)\left( \sqrt{3}+\tan {{20}^{o}} \right)}{\left( 1-\sqrt{3}.\tan {{20}^{o}} \right)\left( 1+\sqrt{3}\tan {{20}^{o}} \right)}\]
Since we know that
\[\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}\]
Therefore, by applying this formula in the above expression, we get,
\[A=\sqrt{3}\left( \tan {{20}^{o}} \right).\left[ \dfrac{{{\left( \sqrt{3} \right)}^{2}}-{{\left( \tan {{20}^{o}} \right)}^{2}}}{{{\left( 1 \right)}^{2}}-{{\left( \sqrt{3}\tan {{20}^{o}} \right)}^{2}}} \right]\]
By simplifying the above equation, we get,
\[\Rightarrow A=\sqrt{3}\left( \tan {{20}^{o}} \right).\left[ \dfrac{3-{{\tan }^{2}}{{20}^{o}}}{1-3{{\tan }^{2}}{{20}^{o}}} \right]\]
Now, we will take \[\tan {{20}^{o}}\] inside the bracket
We get,
\[\Rightarrow A=\sqrt{3}\left[ \dfrac{3\tan {{20}^{o}}-{{\tan }^{3}}{{20}^{o}}}{1-3{{\tan }^{2}}{{20}^{o}}} \right]\]
Since, we know that
\[\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta }=\tan 3\theta \]
Therefore, by applying the above formula, we get,
\[A=\sqrt{3}\left[ \tan 3.\left( {{20}^{o}} \right) \right]\]
We can also write the above expression as
\[A=\sqrt{3}\left[ \tan {{60}^{o}} \right]\]
Since, we know that \[\tan {{60}^{o}}=\sqrt{3}\], therefore by putting the value of \[\tan {{60}^{o}}\] in the above expression we get,
\[A=\sqrt{3}.\sqrt{3}\]
Therefore, A = 3 = RHS
Hence, we proved that the value of \[\tan {{20}^{o}}\tan {{40}^{o}}\tan {{60}^{o}}\tan {{80}^{o}}=3\].
Note: Here by looking at the terms like \[\tan {{20}^{o}},\tan {{40}^{o}}\] and \[\tan {{60}^{o}}\], students often make this mistake of using formulas of double angles that is \[\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }\] which makes the solution lengthy and does not lead to the desired result.
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
