Question

If \[\tan \theta + \tan \left( {\theta + \dfrac{\pi }{3}} \right) + \tan \left( {\theta - \dfrac{\pi }{3}} \right) = k\tan 3\theta \]. Then calculate the value of \[k\].
A. 1
B. 3
C. \[\dfrac{1}{3}\]
D. None of these

Answer 1

Hint: First we will apply the formula \[\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}}\] and \[\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}}\] in the second and third terms of the left side of the given equation. Then simplify the left side expression and apply the formula \[\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}\]. Compare both sides of the equation.

Formula Used:
\[\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}}\]
\[\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}}\]
\[\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}\]

Complete step by step solution:
Given equation is
\[\tan \theta + \tan \left( {\theta + \dfrac{\pi }{3}} \right) + \tan \left( {\theta - \dfrac{\pi }{3}} \right) = k\tan 3\theta \]
Now apply the formulas \[\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}}\] and \[\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}}\]
\[ \Rightarrow \tan \theta + \dfrac{{\tan \theta + \tan \dfrac{\pi }{3}}}{{1 - \tan \theta \tan \dfrac{\pi }{3}}} + \dfrac{{\tan \theta - \tan \dfrac{\pi }{3}}}{{1 + \tan \theta \tan \dfrac{\pi }{3}}} = k\tan 3\theta \]
Now putting \[\tan \dfrac{\pi }{3} = \sqrt 3 \]
\[ \Rightarrow \tan \theta + \dfrac{{\tan \theta + \sqrt 3 }}{{1 - \sqrt 3 \tan \theta }} + \dfrac{{\tan \theta - \sqrt 3 }}{{1 + \sqrt 3 \tan \theta }} = k\tan 3\theta \]
Simplify the left side of the equation
\[ \Rightarrow \tan \theta + \dfrac{{\left( {\tan \theta + \sqrt 3 } \right)\left( {1 + \sqrt 3 \tan \theta } \right) + \left( {\tan \theta - \sqrt 3 } \right)\left( {1 - \sqrt 3 \tan \theta } \right)}}{{\left( {1 - \sqrt 3 \tan \theta } \right)\left( {1 + \sqrt 3 \tan \theta } \right)}} = k\tan 3\theta \]
Apply the formula \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\] in the denominator and simplify the numerator
\[ \Rightarrow \tan \theta + \dfrac{{\tan \theta + \sqrt 3 + \sqrt 3 {{\tan }^2}\theta + 3\tan \theta + \tan \theta - \sqrt 3 - \sqrt 3 {{\tan }^2}\theta + 3\tan \theta }}{{{1^2} - {{\left( {\sqrt 3 \tan \theta } \right)}^2}}} = k\tan 3\theta \]
Cancel out \[\sqrt 3 \] and \[\sqrt 3 {\tan ^2}\theta \]
\[ \Rightarrow \tan \theta + \dfrac{{\tan \theta + 3\tan \theta + \tan \theta + 3\tan \theta }}{{{1^2} - {{\left( {\sqrt 3 \tan \theta } \right)}^2}}} = k\tan 3\theta \]
\[ \Rightarrow \tan \theta + \dfrac{{8\tan \theta }}{{{1^2} - 3{{\tan }^2}\theta }} = k\tan 3\theta \]
Add two terms of left side expression
\[ \Rightarrow \dfrac{{\tan \theta - 3{{\tan }^3}\theta + 8\tan \theta }}{{{1^2} - 3{{\tan }^2}\theta }} = k\tan 3\theta \]
\[ \Rightarrow 3\dfrac{{\left( {3\tan \theta - {{\tan }^3}\theta } \right)}}{{{1^2} - 3{{\tan }^2}\theta }} = k\tan 3\theta \]
Apply the formula \[\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}\] left side expression
\[ \Rightarrow 3\tan 3\theta = k\tan 3\theta \]
\[ \Rightarrow k = 3\]

Hence option B is the correct option.

Note: Students often confused with the formulas \[\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}\] and \[\tan 3\theta = \dfrac{{3\tan \theta + {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}\]. But the correct formula is \[\tan 3\theta = \dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}\].