Question

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

Answer 1

Hint: 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.