91Ó°ÊÓ

Since we are given the polar curve \(r = \sec \theta\), let's first compute the derivative of \(r\) with respect to \(\theta\): $$\frac{dr}{d\theta} = \frac{d}{d\theta}(\sec \theta) = \sec \theta \tan \theta$$

Now, let's compute the Cartesian coordinates \((x,y)\) for a point on the curve using the polar to Cartesian coordinate conversions: $$x = r \cos \theta = (\sec \theta)\cos \theta = 1$$ $$y = r \sin \theta = (\sec \theta)\sin \theta = \tan \theta$$

We will now compute the derivative of \(y\) with respect to \(x\) using the formula mentioned in the analysis: $$\frac{dy}{dx} = \frac{dr}{d\theta}\frac{\sin \theta + r \cos \theta}{r\frac{d\theta}{dx}}$$ Substitute the expressions for \(r\), \(\frac{dr}{d\theta}\), \(x\) and \(y\) into the formula: $$\frac{dy}{dx} = \frac{\sec \theta \tan \theta(\sin \theta + \sec \theta \cos \theta)}{\sec \theta\frac{dx}{d\theta}}$$ Note that \(\frac{dx}{d\theta}\) is the derivative of \(x\), which is a constant and equal to 1, so \(\frac{dx}{d\theta} = 0\). Therefore, we can rewrite the expression for \(\frac{dy}{dx}\) as: $$\frac{dy}{dx} = \sec \theta \tan \theta(\sin \theta + \sec \theta \cos \theta)$$

For a horizontal tangent line, we have \(\frac{dy}{dx} = 0\). From our expression for \(\frac{dy}{dx}\), we get: $$\sec \theta \tan \theta(\sin \theta + \sec \theta \cos \theta) = 0$$ This occurs when \(\tan \theta = 0\), which means \(\theta = n\pi\), where \(n\) is an integer. For a vertical tangent line, we have \(\frac{dy}{dx}\) undefined. This occurs when the denominator of \(\frac{dy}{dx}\) is zero. From our expression for \(\frac{dy}{dx}\), we get: $$\sec \theta \cos \theta = -\sin \theta$$ This occurs when \(\cos \theta = -\frac{\sin \theta}{\sec \theta}\), which means \(\cos^2 \theta = \sin \theta \cos \theta\), so \(\cos \theta = 0\) or \(\sin \theta = \cos \theta\). This implies \(\theta = \frac{\pi}{2} + n\pi\) or \(\theta = \frac{\pi}{4} + n\pi\), where \(n\) is an integer.

For horizontal tangents, we have \(\theta = n\pi\). Thus, plugging these values into the expressions for \(x\) and \(y\), we get the points: $$(x,y) = (1, \tan(n\pi)) = (1, 0)$$ For vertical tangents, we have \(\theta = \frac{\pi}{2} + n\pi\) or \(\theta = \frac{\pi}{4} + n\pi\). Thus, plugging these values into the expressions for \(x\) and \(y\), we get different sets of points: For \(\theta = \frac{\pi}{2} + n\pi\), the points are: $$(x,y) = (1, \tan(\frac{\pi}{2} + n\pi))$$ Note that these points will be undefined, as it causes \(\tan{(\theta)}\) to have either infinity or negative infinity values. For \(\theta = \frac{\pi}{4} + n\pi\), the points are: $$(x,y) = (1, \tan(\frac{\pi}{4} + n\pi))$$ In conclusion, the points on the curve \(r = \sec \theta\) with horizontal tangent lines are \((1, 0)\), and the points with vertical tangent lines have \(\theta\) values of \(\frac{\pi}{4} + n\pi\) or \(\frac{\pi}{2} + n\pi\), with some of these points being undefined due to an infinite \(\tan(\theta)\).