Question

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

Short answer

The value of$\tan \left(\theta \right)+\tan \left(\theta +\pi \right)+\tan \left(\theta +2\pi \right)is9.$

Step-by-step solution

Given information

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

Use periodicity property tanθ+π=tanθ

We get,

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

Conclusion

The value of$\tan \left(\theta \right)+\tan \left(\theta +\pi \right)+\tan \left(\theta +2\pi \right)is9.$