# Prove the following trigonometric equation:

\[\tan {{20}^{o}}.\tan {{40}^{o}}.\tan {{60}^{o}}.\tan {{80}^{o}}=3\]

Answer

Verified

362.7k+ 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.

Last updated date: 24th Sep 2023

â€¢

Total views: 362.7k

â€¢

Views today: 10.62k

Recently Updated Pages

What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Drive an expression for the electric field due to an class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

What is the past tense of read class 10 english CBSE