Question

What is the value of \[\tan \dfrac{{2\pi }}{5} - \tan \dfrac{\pi }{{15}} - \sqrt 3 \tan \dfrac{{2\pi }}{5}\tan \dfrac{\pi }{{15}}\] ?
A. \[ - \sqrt 3 \]
B. \[\dfrac{1}{{\sqrt 3 }}\]
C. \[1\]
D. \[\sqrt 3 \]

Answer 1

Hint: First we will take tangent on both sides of \[\dfrac{{2\pi }}{5} - \dfrac{\pi }{{15}} = \dfrac{\pi }{3}\] . Then we will simplify the equation by using the formula of \[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\] to get the final answer.

Formula Used
\[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\]

Complete step by step solution
We know \[\dfrac{{2\pi }}{5} - \dfrac{\pi }{{15}} = \dfrac{\pi }{3}\]
Taking tan both sides of the equation
\[\tan \left( {\dfrac{{2\pi }}{5} - \dfrac{\pi }{{15}}} \right) = \tan \dfrac{\pi }{3}\]
We know that \[\tan \dfrac{\pi }{3} = \sqrt 3 \]
So \[\tan \left( {\dfrac{{2\pi }}{5} - \dfrac{\pi }{{15}}} \right) = \sqrt 3 \]
Using the formula \[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\] and putting \[A = \dfrac{{2\pi }}{5}\] and \[B = \dfrac{\pi }{{15}}\]
\[\tan \left( {\dfrac{{2\pi }}{5} - \dfrac{\pi }{{15}}} \right) = \dfrac{{\tan \dfrac{{2\pi }}{5} - \tan \dfrac{\pi }{{15}}}}{{1 + \tan \dfrac{{2\pi }}{5}\tan \dfrac{\pi }{{15}}}}\]
So \[\dfrac{{\tan \dfrac{{2\pi }}{5} - \tan \dfrac{\pi }{{15}}}}{{1 + \tan \dfrac{{2\pi }}{5}\tan \dfrac{\pi }{{15}}}} = \sqrt 3 \]
Cross-multiplying the terms
\[\tan \dfrac{{2\pi }}{5} - \tan \dfrac{\pi }{{15}} = \sqrt 3 + \sqrt 3 \tan \dfrac{{2\pi }}{5}\tan \dfrac{\pi }{{15}}\]
Rearranging the terms
\[\tan \dfrac{{2\pi }}{5} - \tan \dfrac{\pi }{{15}} - \sqrt 3 \tan \dfrac{{2\pi }}{5}\tan \dfrac{\pi }{{15}} = \sqrt 3 \]
Hence option D is correct.

Additional information: We know that \[\cos 90^{\circ} = 0\] and \[\tan 90^{\circ} = \dfrac {\sin 90^{\circ}}{\cos 90^{\circ}}\]. If the denominator of a rational number is zero then the rational number is known as undefined. Thus the value of \[\tan 90^{\circ}\] is undefine.


Note: Students must not get confused between the formula of sum and difference of angles between tan; the formulas are as follows –
\[\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}\]
\[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\]