Question

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

Answer 1

Hint: In the given question, we have been asked to find the value of a trigonometric function. Now, the argument of the given trigonometric function is not in the range of the known values of the trigonometric functions as given in the standard table, in which values lie from \[0\] to \[\pi /2\]. But we can calculate that by using the formula of periodicity of the given trigonometric function and then solving it.

Formula Used:
We are going to use the formula of periodicity of tangent function, which is:
\[\tan \left( {\pi + \theta } \right) = \tan \left( \theta \right)\]

Complete step-by-step answer:
Here, we have to calculate the value of \[\tan \left( {\dfrac{{7\pi }}{6}} \right)\].
Now, we know that the periodicity of tangent function is \[\pi \].
Hence, \[\tan \left( {\dfrac{{7\pi }}{6}} \right) = \tan \left( {\pi + \dfrac{\pi }{6}} \right) = \tan \left( {\dfrac{\pi }{6}} \right)\]
Now, we know that
\[\tan \left( {\dfrac{\pi }{6}} \right) = \dfrac{1}{{\sqrt 3 }}\]
Hence, \[\tan \left( {\dfrac{{7\pi }}{6}} \right) = \dfrac{1}{{\sqrt 3 }}\]

Additional Information:
In the given question, we applied the concept of periodicity of the tangent function, so it is necessary that we know the periodicity of each trigonometric function. Periodicity of sine, cosine, cosecant and secant is \[2\pi \] while the periodicity of tangent and cotangent is \[\pi \].

Note: We just need to remember the periodicity of the trigonometric functions, like here, the tangent function has the periodicity of \[\pi \], i.e every \[\tan \] value repeats after this interval, so, when the angle is more than \[180^\circ \] or \[\pi \], we convert it to the smaller value and solve for the answer using the smaller value.