$$ 0 = x\cos\theta\cos\phi+y\sin\theta\cos\phi+z\sin\phi$$
Here's what I've tried doing.
$$\begin{align} x\cos\theta\cos\phi+y\sin\theta\cos\phi & = -z\sin\phi \\ x\cos\theta+y\sin\theta & = -z\tan\phi \\ x\cos\theta+y\sqrt{1-\cos^2\theta} & = -z\tan\phi \\ y\sqrt{1-\cos^2\theta} & = -x\cos\theta-z\tan\phi \\ y^2-y^2\cos^2\theta & = x^2\cos^2\theta-z^2\tan^2\phi+2xz\cos\theta\tan\phi \\ 0 & = (x^2+y^2)\cos^2\theta+2xz\tan\phi\cos\theta-y^2-z^2\tan^2\phi \end{align}$$ $$\begin{align} \cos\theta & = \frac{-2xz\tan\phi\pm\sqrt{4x^2z^2\tan^2\phi-4(x^2+y^2)(-y^2-z^2\tan^2\phi)}}{2(x^2+y^2)} \\ \cos\theta & = \frac{-2xz\tan\phi\pm2\sqrt{x^2z^2\tan^2\phi+y^2(x^2+y^2)-x^2z^2\tan^2\phi-y^2z^2\tan^2\phi}}{2(x^2+y^2)} \\ \cos\theta & = \frac{-xz\tan\phi\pm y\sqrt{x^2+y^2-z^2\tan^2\phi}}{x^2+y^2} \\ \end{align}$$
However, the $\pm$ is throwing me off. I'm assuming that either the $+$ or $-$ gives a complex solution or something, but I'm not sure what.
HINT :
$$ x\cos\theta+y\sin\theta=r\cos(\theta-\alpha), $$ where $r^2=x^2+y^2$ and $\tan\alpha=\dfrac{y}{x}$. You may refer to this link.
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$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$