Let $$P(x, y, z)$$ be a point in the first octant, whose projection in the $$xy$$-plane is the point $$Q$$. Let $$OP = \gamma$$; the angle between $$OQ$$ and the positive $$x$$-axis be $$\theta$$; and the angle between $$OP$$ and the positive $$z$$-axis be $$\phi$$, where $$O$$ is the origin. Then the distance of $$P$$ from the $$x$$-axis is

We wish to find the distance of point $$P$$ from the x-axis using spherical-like coordinates.

Since $$OP = \gamma$$, we let $$\phi$$ denote the angle between $$OP$$ and the positive $$z$$-axis and $$\theta$$ denote the angle between the projection $$OQ$$ onto the $$xy$$-plane and the positive $$x$$-axis. This gives the coordinate expressions $$x = \gamma\sin\phi\cos\theta,\quad y = \gamma\sin\phi\sin\theta,\quad z = \gamma\cos\phi.$$

Since the distance from the $$x$$-axis is given by $$d = \sqrt{y^2 + z^2},$$ substituting the expressions for $$y$$ and $$z$$ yields $$d = \sqrt{\gamma^2\sin^2\phi\sin^2\theta + \gamma^2\cos^2\phi} = \gamma\sqrt{\sin^2\phi\sin^2\theta + \cos^2\phi}.$$

This expression can be simplified by noting that $$\sin^2\phi\sin^2\theta + \cos^2\phi = 1 - \sin^2\phi + \sin^2\phi\sin^2\theta = 1 - \sin^2\phi\bigl(1 - \sin^2\theta\bigr) = 1 - \sin^2\phi\cos^2\theta,$$ so that $$d = \gamma\sqrt{1 - \sin^2\phi\cos^2\theta}.$$

The correct answer is Option (1): $$\gamma\sqrt{1 - \sin^2\phi\cos^2\theta}.$$