Question

How do you evaluate \[\tan (\dfrac{{2\pi }}{3})\]?

Answer 1

Hint:
Real functions that relate any angle of a right-angled triangle to the ratio of any two of its sides are Trigonometric functions. We can also use geometric definitions to evaluate trigonometric values. Here, it’s important that we know the sine of theta is the ratio of the opposite side to the hypotenuse and the cosine of theta is the ratio of the adjacent side (base) to the hypotenuse. Also, it is known that the ratio of the sine of theta to the cosine of theta is cot of theta.

Complete step by step:
According to the given data, we need to evaluate \[\tan (\dfrac{{2\pi }}{3})\]
If in a right angled triangle \[\theta \] represents one of its acute angle then by definition we can write
\[\sin \theta = \dfrac{{Opposite}}{{Hypotenuse}}\]
Also,\[\cos \theta = \dfrac{{Base}}{{Hypotenuse}}\]
Thereafter we know that,
\[\tan \theta = \dfrac{1}{{\cot \theta }} = \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{Opposite}}{{Base}}\]
Therefore,
\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
In the given case, \[\theta = \dfrac{{2\pi }}{3}\]
\[\therefore \tan (\dfrac{{2\pi }}{3}) = \dfrac{{\sin (\dfrac{{2\pi }}{3})}}{{\cos (\dfrac{{2\pi }}{3})}}\]

According to the unit circle, it is known to us that,
\[\cos (\dfrac{{2\pi }}{3}) = - \dfrac{1}{2}\] and, \[\sin (\dfrac{{2\pi }}{3}) = \dfrac{{\sqrt 3 }}{2}\].

When we substitute the values in the expression we get,
\[ \Rightarrow \tan (\dfrac{{2\pi }}{3}) = \dfrac{{\sin (\dfrac{{2\pi }}{3})}}{{\cos (\dfrac{{2\pi }}{3})}} = \dfrac{{\dfrac{{\sqrt 3 }}{2}}}{{ - \dfrac{1}{2}}}\]

\[ \Rightarrow \tan (\dfrac{{2\pi }}{3}) = - \sqrt 3 \]

Hence, the value of \[\tan (\dfrac{{2\pi }}{3})\] is \[ - \sqrt 3 \].

Note:
One should be careful while evaluating trigonometric values and rearranging the terms to convert from one function to the other. The widely used ones are sin, cos, and tan. While the rest can be referred to as the reciprocal of the others, i.e., cosec, sec, and cot respectively. If in a right-angled triangle θ represents one of its acute angles then, \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\].