Q.91 - Precalculus Enhanced with Graphing Utilities

Question

if $\tan \left(\theta \right)=3$ then find $\tan \left(\theta \right)+\tan \left(\theta +\pi \right)+\tan \left(\theta +2\pi \right)$

Step-by-Step Solution

Answer

The value of $\tan \left(\theta \right)+\tan \left(\theta +\pi \right)+\tan \left(\theta +2\pi \right) is 9.$

Step 1: Given information

We are given $\tan \left(\theta \right)=3$

Step 2: Use periodicity property tan θ + π = tan θ

We get,

$$\begin{gathered} \tan \left(\theta \right)+\tan \left(\theta +\pi \right)+\tan \left(\theta +2\pi \right) \\ \tan \left(\theta \right)+\tan \left(\theta \right)+\tan \left(\theta +\pi \right) \\ \tan \left(\theta \right)+\tan \left(\theta \right)+\tan \left(\theta \right) \\ 3\tan \left(\theta \right) \\ =3\left(3\right) \\ =9 \end{gathered}$$

Step 3: Conclusion

The value of $\tan \left(\theta \right)+\tan \left(\theta +\pi \right)+\tan \left(\theta +2\pi \right) is 9.$