The spherical coordinates of $(-3, 4, -12)$ are $(\rho, \theta, \phi)$. Find $\tan \theta + \tan \phi$.

Question

The spherical coordinates of $(-3, 4, -12)$ are $(\rho, \theta, \phi)$. Find $\tan (\theta )+ \tan( \phi)$.

I've already tried converting the Cartesian coordinates provided into spherical coordinates:

I got $\rho= 13$, $\theta = \arccos(-12/13)$, and $\phi = \arctan(-4/3)$

I'm not sure how to add theta and phi together, I'm not sure if I converted them correctly at all.

Answer 1

Draw a 'triangle' with $\cos(\theta)=-12/13$ (the one side will be negative, but this is fine); what is $\tan(\theta)$ in this triangle? Also, if $\phi = \arctan(-4/3)$, what is $\tan(\phi)$?

Answer 2

If $$x=r \sin \theta \cos \phi$$ $$y=r \sin \theta \sin \phi$$ $$z = r \cos \theta$$ Where $\theta \in [0, \pi] $ is angle between axis $z$ and interval ended in point in question and $\phi \in [0, 2 \pi)$ is angle between $x$ axis and projection of point on $xy$ plane.Then we have $$r=\sqrt{x^2+y^2+z^2}$$ $$\theta= \arctan \frac{\sqrt{x^2+y^2}}{z} = \arccos \frac{z}{\sqrt{x^2+y^2+z^2}}$$ $$\phi = \arctan \frac{y}{x}$$