Question

How do you evaluate $\tan (210)?$

Answer 1

Hint:The trigonometric function $\tan \theta$ is defined for all real numbers $\theta $. We can write it in terms of $\sin \theta $ and $\cos \theta $ as $\dfrac{\sin \theta }{\cos \theta }$, where $\cos \theta \ne 0$.
The function $\tan \theta $ is undefined for $\theta =...,-\dfrac{5\pi }{2},-\dfrac{3\pi }{2},-\dfrac{\pi }{2},\dfrac{\pi }{2},\dfrac{3\pi }{2},\dfrac{5\pi }{2},...$
Some properties will help to evaluate $\tan (210)$
$\Rightarrow \pi $(in radian)$={{180}^{0}}$
$\Rightarrow \tan (\pi +\theta )=\tan \theta $.
$\Rightarrow \tan (-\theta )=-\tan \theta $.
$\Rightarrow $$\tan \left( \dfrac{\pi }{2}-\theta \right)=\cot \theta $.
$\Rightarrow $$\tan \left( \dfrac{\pi }{2}+\theta \right)=\tan \theta $.
$\Rightarrow $$\tan (\theta +\varphi )=\dfrac{\tan \theta +\tan \varphi }{1-\tan \theta \times \tan \varphi }$.

Complete step by step solution:
To evaluate $\tan (210)$ we will do two factors $210$ in terms of $(\pi +\theta )$.
Therefore, $210$ can be written as in term of $(\pi +\theta )$is $(180+30)$
$\Rightarrow \tan (210)=\tan (\pi +30)$
$\Rightarrow \tan (\pi +30)=\tan (30)$
$\Rightarrow \tan (30)$
$\Rightarrow \dfrac{1}{\sqrt{3}}$
$\tan (210)=\dfrac{1}{\sqrt{3}}$.
We can rewrite this solution $\dfrac{\sqrt{3}}{3}$ by use of rationalization of the denominator as
$\Rightarrow \dfrac{1}{\sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}}$
$\Rightarrow \dfrac{\sqrt{3}}{3}$.
The value of $\tan (210)$both is equal

Hence the value $\tan (210)$ will be $\dfrac{1}{\sqrt{3}}$ or $\dfrac{\sqrt{3}}{3}$.

Note: The angle 210 will be in the third quadrant.
The tangent function is undefined when the function \[\cos ine\] will be zero.
The domain of the function $\tan \theta $ is real numbers.
And the range of the function is real numbers.