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

Dec 16, 2019

#1
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The spherical coordinates of (-3, 4, -12) are $$(\rho, \theta, \varphi)$$.
Find $$\tan \theta + \tan \varphi$$.

I assume: the spherical coordinates are $$(\text{radius}~ \rho, \text{inclination}~ \theta, \text{azimuth}~ \varphi)$$

$$\begin{array}{|lrcrl|} \hline (1) & x &=& -3 &= \rho \sin(\theta) \cos(\varphi) \\ (2) & y &=& 4 &= \rho \sin(\theta) \sin(\varphi) \\ (3) & z &=& -12 &= \rho \cos(\theta) \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline \dfrac{(2)}{(1)}: & \dfrac{\rho \sin(\theta) \sin(\varphi)}{\rho \sin(\theta) \cos(\varphi)} &=& \dfrac{4}{-3} \\ & \mathbf{\tan(\varphi)} &=& \mathbf{-\dfrac{4}{3}} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline \dfrac{(2)}{(3)}: & \dfrac{\rho \sin(\theta) \sin(\varphi)} {\rho \cos(\theta)} &=& \dfrac{4}{ -12} \\ & \tan(\theta)\sin(\varphi) &=& -\dfrac{1}{3} \quad | \quad \text{square both sides} \\ & \mathbf{\tan^2(\theta)\sin^2(\varphi)} &=& \mathbf{\dfrac{1}{9}} \\ \hline \dfrac{(1)}{(3)}: & \dfrac{\rho \sin(\theta) \cos(\varphi)} {\rho \cos(\theta)} &=& \dfrac{-3}{ -12} \\ & \tan(\theta)\cos(\varphi) &=& \dfrac{1}{4} \quad | \quad \text{square both sides} \\ & \mathbf{\tan^2(\theta)\cos^2(\varphi)} &=& \mathbf{\dfrac{1}{16}} \\ \hline & \tan^2(\theta)\sin^2(\varphi)+\tan^2(\theta)\cos^2(\varphi) &=& \dfrac{1}{9}+\dfrac{1}{16} \\ & \tan^2(\theta)\left(\underbrace{ \sin^2(\varphi)+ \cos^2(\varphi)}_{=1} \right) &=& \dfrac{25}{9*16} \\ & \tan^2(\theta) &=& \dfrac{25}{9*16} \\ & \tan(\theta) &=& \dfrac{5}{3*4} \\ & \mathbf{\tan(\theta)} &=& \mathbf{\dfrac{5}{12}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \tan(\theta) + \tan(\varphi) &=& \dfrac{5}{12} -\dfrac{4}{3} \\\\ &=& \dfrac{15-48}{36} \\\\ &=& \dfrac{-33}{36} \\\\ \mathbf{\tan(\theta) + \tan(\varphi)} &=& \mathbf{-\dfrac{11}{12}} \\ \hline \end{array}$$

Dec 16, 2019