Find the equation of a straight line joining the pair of points $\left(a \cos \theta, b \sin \theta\right)$ and $\left(a \cos \phi, b \sin \phi\right)$.
Let $\;$ $A \left(x_1, y_1\right) = \left(a \cos \theta, b \sin \theta\right)$, $\;$ $B \left(x_2, y_2\right) = \left(a \cos \phi, b \sin \phi\right)$
Slope of the required line $= m = \dfrac{y_2 - y_1}{x_2 - x_1}$
i.e. $\;$ $m = \dfrac{b \sin \phi - b \sin \theta}{a \cos \phi - a \cos \theta}$
i.e. $\;$ $m = \dfrac{b \left(\sin \phi - \sin \theta\right)}{a \left(\cos \phi - \cos \theta\right)}$
$\left[\text{Note: } \sin C - \sin D = 2 \cos \left(\dfrac{C + D}{2}\right) \sin \left(\dfrac{C - D}{2}\right) \right.$
$\left. \cos C - \cos D = - 2 \sin \left(\dfrac{C + D}{2}\right) \sin \left(\dfrac{C - D}{2}\right) \right]$
i.e. $\;$ $m = \dfrac{2b \cos \left(\dfrac{\theta + \phi}{2}\right) \sin \left(\dfrac{\phi - \theta}{2}\right)}{-2a \sin \left(\dfrac{\theta + \phi}{2}\right) \sin \left(\dfrac{\phi - \theta}{2}\right)}$
i.e. $\;$ $m = \dfrac{-b}{a} \times \dfrac{\cos \left(\dfrac{\theta + \phi}{2}\right)}{\sin \left(\dfrac{\theta + \phi}{2}\right)}$
Equation of the required line is of the form: $\;$ $y - y_1 = m \left(x - x_1\right)$
i.e. $\;$ $y - b \sin \theta = \dfrac{-b}{a} \times \dfrac{\cos \left(\dfrac{\theta + \phi}{2}\right)}{\sin \left(\dfrac{\theta + \phi}{2}\right)} \times \left(x - a \cos \theta\right)$
i.e. $\;$ $y \sin \left(\dfrac{\theta + \phi}{2}\right)a - ab \sin \theta \sin \left(\dfrac{\theta + \phi}{2}\right)$
$\hspace{1.5cm}$ $ = - b \cos \left(\dfrac{\theta + \phi}{2}\right)x + ab \cos \theta \cos \left(\dfrac{\theta + \phi}{2}\right)$
i.e. $\;$ $x \cos \left(\dfrac{\theta + \phi}{2}\right)b + y \sin \left(\dfrac{\theta + \phi}{2}\right)a$
$\hspace{1.5cm}$ $ = ab \left[\sin \theta \sin \left(\dfrac{\theta + \phi}{2}\right) + \cos \theta \cos \left(\dfrac{\theta + \phi}{2}\right)\right]$
$\left[\text{Note: } \cos \left(A - B\right) = \cos A \cos B + \sin A \sin B\right]$
i.e. $\;$ $\dfrac{x}{a} \cos \left(\dfrac{\theta + \phi}{2}\right) + \dfrac{y}{b} \sin \left(\dfrac{\theta + \phi}{2}\right) = \cos \left(\theta - \dfrac{\theta + \phi}{2}\right)$
i.e. $\;$ $\dfrac{x}{a} \cos \left(\dfrac{\theta + \phi}{2}\right) + \dfrac{y}{b} \sin \left(\dfrac{\theta + \phi}{2}\right) = \cos \left(\dfrac{\theta - \phi}{2}\right)$