Explanation

(a) \(\frac{1}{1 - \cos \theta} + \frac{1}{1 + \cos \theta}\)

= \(\frac{(1 + \cos \theta) + (1 - \cos \theta)}{1 - \cos^{2} \theta}\)

= \(\frac{2}{1 - \cos^{2} \theta}\)

= \(\frac{2}{\sin^{2} \theta}\)

(b) \(y = x^{2} (x - 3)\)

\(\frac{\mathrm d y}{\mathrm d x} = x^{2} + 2x(x - 3)\)

= \(x^{2} + 2x^{2} - 6x = 3x^{2} - 6x\)

At stationary point, \(\frac{\mathrm d y}{\mathrm d x} = 0\)

\(3x^{2} - 6x = 0\)

\(3x(x - 2) = 0 \implies x = \text{0 or 2}\)

If x = 0, \(y = 0^{2}(0 - 3) = 0\)

If x = 2, \(y = 2^{2}(2 - 3) = -4\)

The points are (0, 0) and (2, -4).

Gradient = \(\frac{-4 - 0}{2 - 0} = \frac{-4}{2} = -2\)

Equation : \(\frac{y - (-4)}{x - 2} = -2\)

\(\frac{y + 4}{x - 2} = -2 \implies y + 4 = -2(x - 2)\)

\(y + 4 = 4 - 2x \implies y + 2x = 0\)

\(y = -2x\)

Distance : \(\sqrt{(2 - 0)^{2} + (-4 - 0)^{2}}\)

= \(\sqrt{4 + 16} = \sqrt{20}\)

= \(2\sqrt{5} units\)

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