Question

Prove that $\tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ = 3$.

Answer 1

Hint: First, break $\tan 80^\circ $ and $\tan 20^\circ $ in terms of $\tan 60^\circ $ and $\tan 20^\circ $ by using a basic formula $\tan \left( {A \pm B} \right) = \dfrac{{\tan A \pm \tan B}}{{1 \mp \tan A\tan B}}$. After that multiply both the equations and simplify it. Again, multiply by $\tan 20^\circ $ and simplify it. The equation will be in the form $\dfrac{{3\tan A - {{\tan }^3}A}}{{1 - 3{{\tan }^2}A}}$ which can be replaced by $\tan 3A$. Now, the term \[\tan 20^\circ \tan 40^\circ \tan 80^\circ \] will be equal to $\tan 60^\circ $. So, the term\[\tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ \] is equal to ${\left( {\tan 60^\circ } \right)^2}$ whose value is 3.

Formula used:
$\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
$\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
$\tan 3A = \dfrac{{3\tan A - {{\tan }^3}A}}{{1 - 3{{\tan }^2}A}}$

Complete step-by-step answer:
To prove:- $\tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ = 3$
Break-in terms of $\tan 20^\circ $ and $\tan 60^\circ $ by using $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$,
\[\tan 80^\circ = \dfrac{{\tan 60^\circ + \tan 20^\circ }}{{1 - \tan 60^\circ \tan 20^\circ }}\]
Substitute $\tan 60^\circ = \sqrt 3 $ we get,
$\tan 80^\circ = \dfrac{{\sqrt 3 + \tan 20^\circ }}{{1 - \sqrt 3 \tan 20^\circ }}$ …..(1)
Now, break-in $\tan 40^\circ $ in terms of $\tan 20^\circ $ and $\tan 60^\circ $ by using $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$,
\[\tan 40^\circ = \dfrac{{\tan 60^\circ - \tan 20^\circ }}{{1 + \tan 60^\circ \tan 20^\circ }}\]
Substitute $\tan 60^\circ = \sqrt 3 $ we get,
$\tan 40^\circ = \dfrac{{\sqrt 3 - \tan 20^\circ }}{{1 + \sqrt 3 \tan 20^\circ }}$ …..(2)
Now multiply equation (1) and (2),
$\tan 40^\circ \tan 80^\circ = \dfrac{{\sqrt 3 - \tan 20^\circ }}{{1 + \sqrt 3 \tan 20^\circ }} \times \dfrac{{\sqrt 3 + \tan 20^\circ }}{{1 - \sqrt 3 \tan 20^\circ }}$
Multiply the values of the numerator and denominator,
$\tan 40^\circ \tan 80^\circ = \dfrac{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\tan 20^\circ } \right)}^2}}}{{{{\left( 1 \right)}^2} - {{\left( {\sqrt 3 \tan 20^\circ } \right)}^2}}}$
Open brackets and square the terms inside corresponding brackets,
$\tan 40^\circ \tan 80^\circ = \dfrac{{3 - {{\tan }^2}20^\circ }}{{1 - 3{{\tan }^2}20^\circ }}$
Multiply both sides by $\tan 20^\circ $,
$\tan 20^\circ \tan 40^\circ \tan 80^\circ = \tan 20^\circ \times \dfrac{{3 - {{\tan }^2}20^\circ }}{{1 - 3{{\tan }^2}20^\circ }}$
Multiply $\tan 20^\circ $ in the numerator,
\[\tan 20^\circ \tan 40^\circ \tan 80^\circ = \dfrac{{3\tan 20^\circ - {{\tan }^3}20^\circ }}{{1 - 3{{\tan }^2}20^\circ }}\]
As we know that $\tan 3A = \dfrac{{3\tan A - {{\tan }^3}A}}{{1 - 3{{\tan }^2}A}}$. Then, the term in the right-hand side will be replaced,
\[\tan 20^\circ \tan 40^\circ \tan 80^\circ = \tan \left( {3 \times 20^\circ } \right)\]
Multiply both sides by $\tan 60^\circ $ we get,
\[\tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ = {\left( {\tan 60^\circ } \right)^2}\]
Substitute the value of \[\tan 60^\circ \] on the right side of the equation,
\[\tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ = {\left( {\sqrt 3 } \right)^2}\]
Open the bracket and square the term on the right side,
\[\tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ = 3\]
Hence, it is proved.

Note: The students are likely to make mistakes by converting tan in form of sin and cos. It will make the problem complicated and lengthy.
Trigonometry is concerned with specific functions of angles and their application to calculations. There are six functions of an angle commonly used in trigonometry. Their names and abbreviations are sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec), and cosecant (cosec).