Question

How do you find the exact value of $\tan \left( {\dfrac{{7\pi }}{6}} \right)$ ?

Answer 1

Hint : In the given problem, we are required to find the tangent of a given angle using some simple and basic trigonometric compound angle formulae and trigonometric identities such as $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ . Such questions require basic knowledge of compound angle formulae and their applications in this type of questions.

Complete step-by-step answer :
Consider a unit circle (a circle of radius of 1 unit centered at origin). We need to find out the value of $\cos \dfrac{{53\pi }}{6}$ using the unit circle.
So, we have, $\tan \left( {\dfrac{{7\pi }}{6}} \right)$ .
We know the compound angle formula for tangent as $\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ . So, splitting the angle of the trigonometric ratio into two parts, we get,
$\Rightarrow \tan \left( {\dfrac{{7\pi }}{6}} \right) = \tan \left( {\pi + \dfrac{\pi }{6}} \right)$
So, using the tangent compound angle formula, we get,
$\Rightarrow \tan \left( {\dfrac{{7\pi }}{6}} \right) = \dfrac{{\tan \left( \pi \right) + \tan \left( {\dfrac{\pi }{6}} \right)}}{{1 - \tan \left( \pi \right)\tan \left( {\dfrac{\pi }{6}} \right)}}$
Now, we know that the value of the tangent trigonometric function at angle $\pi$ radians is zero. So, we get,
$\Rightarrow \tan \left( {\dfrac{{7\pi }}{6}} \right) = \dfrac{{0 + \tan \left( {\dfrac{\pi }{6}} \right)}}{{1 - \left( 0 \right)\tan \left( {\dfrac{\pi }{6}} \right)}}$
Simplifying the expression further, we get,
$\Rightarrow \tan \left( {\dfrac{{7\pi }}{6}} \right) = \dfrac{{\tan \left( {\dfrac{\pi }{6}} \right)}}{{1 - 0}}$
We know that the value of $\tan \left( {\dfrac{\pi }{6}} \right)$ is $\dfrac{1}{{\sqrt 3 }}$ . So, we get,
$\Rightarrow \tan \left( {\dfrac{{7\pi }}{6}} \right) = \dfrac{1}{{\sqrt 3 }}$
Hence, the value of $\tan \left( {\dfrac{{7\pi }}{6}} \right)$ is $\left( {\dfrac{1}{{\sqrt 3 }}} \right)$ .
So, the correct answer is “$\left( {\dfrac{1}{{\sqrt 3 }}} \right)$”.

Note : Periodic Function is a function that repeats its value after a certain interval. The given problem can also be solved using the periodicity of the tangent function. We know that the tangent function has $\pi$ radians as its fundamental period. So, the value of $\tan \left( {\dfrac{{7\pi }}{6}} \right)$ is same as value of $\tan \left( {\dfrac{{7\pi }}{6} - \pi } \right)$ . For a real number $T > 0$ , $f\left( {x + T} \right) = f\left( x \right)$ for all x, if T is the smallest positive real number such that $f\left( {x + T} \right) = f\left( x \right)$ for all x, then T is called the fundamental period. Take care of the calculative steps in order to be sure of the final answer.