Question

How do you evaluate $\tan (-30°)$?

Answer 1

Finding the required value:

Since, $\tan \left(\theta \right)=\frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}$

$$\begin{gathered} \tan (-30°)=\frac{\sin (-30°)}{\cos (-30°)} \\ \tan (-30°)=\frac{-\sin (30°)}{\cos (30°)} \end{gathered}$$ $\left[\because \sin (-\mathrm{\theta })=-\mathrm{sin\theta },\cos (-\mathrm{\theta })=\mathrm{cos\theta }\right]$

$$\begin{gathered} \tan (-30°)=\frac{-\sin (30°)}{\cos (30°)} \\ \tan (-30°)=\frac{-{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{3}}{2}}} \\ \tan (-30°)=-\frac{1}{\sqrt{3}} \end{gathered}$$ $\left[\sin (30°)=\frac{1}{2},\cos (30°)=\frac{1}{\sqrt{3}}\right]$

Hence, the value of $\tan (-30°)$ is $-\frac{1}{\sqrt{3}}$.