Question

How do you calculate $\tan \left( {\dfrac{{7\pi }}{6}} \right)?$

Answer 1

As $\tan \left( {\dfrac{{7\pi }}{6}} \right)$ is not the standard so we can convert this into one, by breaking down the angle into parts, on will be $\dfrac{\pi }{6}$ and other will be $\pi$ only and also you can see that $\dfrac{\pi }{6} + \pi = \dfrac{{7\pi }}{6}$ and then we can use trigonometric results to solve this question.

Complete step-by-step solution:
As said, $\tan \left( {\dfrac{{7\pi }}{6}} \right)$ can be written as $\tan \left( {\dfrac{\pi }{6} + \pi } \right)$ , So our equation will become,
$\Rightarrow \tan \left( {\dfrac{{7\pi }}{6}} \right) = \tan \left( {\dfrac{\pi }{6} + \pi } \right)$
Now using the trigonometric identity for tan (A+B) which says that,
$\Rightarrow \tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - tanA\tan B}}$
So, $\tan (\dfrac{\pi }{6} + \pi )$ can be written as follows,
$\tan (\dfrac{\pi }{6} + \pi ) = \dfrac{{\tan \dfrac{\pi }{6} + \tan \pi }}{{1 - tan\dfrac{\pi }{6}\tan \pi }}$
From the standard results, we have that the value of $\tan \dfrac{\pi }{6}$ is equal to $\dfrac{1}{{\sqrt 3 }}$ and the value of $\tan \pi$ is equal to $0$ , so putting these values in the above equation will give us,
$\Rightarrow \tan (\dfrac{\pi }{6} + \pi ) = \dfrac{{\dfrac{1}{{\sqrt 3 }} + 0}}{{1 - \dfrac{1}{{\sqrt 3 }} \times 0}}$
Solving the above equation will give us,
$\Rightarrow \tan (\dfrac{\pi }{6} + \pi ) = \dfrac{{\dfrac{1}{{\sqrt 3 }}}}{{1 - 0}}$
Which means that the value of $\tan (\dfrac{\pi }{6} + \pi )$ is
$\Rightarrow \tan (\dfrac{\pi }{6} + \pi ) = \dfrac{1}{{\sqrt 3 }}$
Back substituting $\tan (\dfrac{\pi }{6} + \pi )$ as $\tan (\dfrac{{7\pi }}{6})$ ,
$\Rightarrow \tan (\dfrac{{7\pi }}{6}) = \dfrac{1}{{\sqrt 3 }}$
This is our required answer.

Thus the corrected answer is $\tan (\dfrac{{7\pi }}{6}) = \dfrac{1}{{\sqrt 3 }}$

Note: The angle can also be written in the form of difference of two angles where we would have used the identity of tan (A-B) which is also a standard identity. Here one should learn the results of both tan (A+B) and tan (A-B) as both are very important and one should also learn the value of trigonometric ratios at standard angles.