Q. 86 - Precalculus Enhanced with Graphing Utilities

Question

Find the exact value of:$\tan 60°+\tan 150°$

Step-by-Step Solution

Answer

The value of $\tan 60°+\tan 150°=\frac{2\sqrt{3}}{3}$

Step 1. Given information

The given value $=\tan 60°+\tan 150°$

Step 2. Find the exact value of the given function by using unit circle

The unit circle 

Step 3. Finding the value of tan 60 ° + tan 150 °

Recall that the unit circle gives $(x,y)=(\cos \alpha ,\sin \alpha )$ for all the common angles:
Consider $\tan 60°$
Let $\alpha =60°$
From the above unit circle,
The angle $\alpha =60°$ intersects the unit circle at $\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$ as shown in the above figure.
The pair of coordinates corresponding to $60°$ is $\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$ ,
Here the first coordinate gives $\cos 60°$ and second coordinate gives $\sin 60°$ .
Therefore,
$\cos 60°=\frac{1}{2},\sin 60°=\frac{\sqrt{3}}{2}$

since $\tan \theta =\frac{\sin \theta }{\cos \theta }$

Replace $\theta$ by $60°$ then

$$\begin{gathered} \tan 60°=\frac{\sin 60°}{\cos 60°} \\ =\frac{{\displaystyle \frac{\sqrt{3}}{2}}}{{\displaystyle \frac{1}{2}}} [\sin ce \cos 60°=\frac{1}{2},\sin 60°=\frac{\sqrt{3}}{2}] \\ =\frac{\sqrt{3}}{1} \\ =\sqrt{3} \end{gathered}$$

Therefore,the value of $\tan 60° is \sqrt{3}$

Consider $\tan 150°$
Let $\alpha =150°$
From the above unit circle,
The angle $\alpha =150°$ intersects the unit circle at $\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)$ as shown in the above figure.
The pair of coordinates corresponding to $150°$ is $\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)$,
Here the first coordinate gives $\cos 150°$ and second coordinate gives $\sin 150°$ .
Therefore, 

$\cos 150°=-\frac{\sqrt{3}}{2},\sin 150°=\frac{1}{2}$

since $\tan \theta =\frac{\sin \theta }{\cos \theta }$

Replace $\theta$ by $150°$then

$$\begin{gathered} \tan 150°=\frac{\sin 150°}{\cos 150°} \\ =\frac{{\displaystyle \frac{1}{2}}}{-{\displaystyle \frac{\sqrt{3}}{2}}} [\sin ce \cos 150°=-\frac{\sqrt{3}}{2},\sin 150°=\frac{1}{2}] \\ =\frac{-1}{\sqrt{3}} \end{gathered}$$

Therefore,the value of $\tan 150° is \frac{-1}{\sqrt{3}}$

Finally,add these two exact functional values to obtain the required solution

Therefore,

$$\begin{gathered} \tan 60°+\tan 150°=\sqrt{3}+\frac{-1}{\sqrt{3}} \\ =\sqrt{3}+\left(\frac{-1}{\sqrt{3}}\right) \\ =\frac{3\sqrt{3}-\sqrt{3}}{3} \\ =\frac{2\sqrt{3}}{3} \end{gathered}$$