A geometry problem by Md Zuhair

$\begin{aligned} \frac x{1-\sin \theta} + \frac y{1+\sin \theta} & = \frac 1{\cos \theta} \\ \frac {x(1+\sin \theta) + y(1-\sin \theta)}{(1+\sin \theta) (1-\sin \theta)} & = \frac 1{\cos \theta} \\ \frac {x(1+\sin \theta) + y(1-\sin \theta)}{1-\sin^2 \theta} & = \frac 1{\cos \theta} \\ \frac {x(1+\sin \theta) + y(1-\sin \theta)}{\cos^2 \theta} & = \frac 1{\cos \theta} \\ x(1+\sin \theta) + y(1-\sin \theta) & = \cos \theta & \small \color{#3D99F6} \implies \text{The graph is a straight line for a }\theta. \\ 0 + {\color{#3D99F6}y_0}(1-\sin \theta) & = \cos \theta & \small \color{#3D99F6} \text{where }y_0 \text{ is the }y \text{-axis intercept.} \theta \\ \implies y_0 & = \frac {\cos \theta}{1-\sin \theta} \\ {\color{#3D99F6}x_0}(1+\sin \theta) + 0 & = \cos \theta & \small \color{#3D99F6} \text{where }x_0 \text{ is the }x \text{-axis intercept.} \theta \\ \implies x_0 & = \frac {\cos \theta}{1+\sin \theta} \end{aligned}$

The area of between $x$ -axis, $y$ -axis and the graph is an area of a right-angled triangle:

$\begin{aligned} A & = \frac 12 x_0 y_0 \\ & = \frac 12 \cdot \frac {\cos^2 \theta}{(1+\sin \theta)(1-\sin \theta)} \\ & = \frac {\cos^2 \theta}{2(1-\sin^2 \theta)} \\ & = \frac {\cos^2 \theta}{2\cos^2 \theta} \\ & = \frac 12 = \boxed{0.5} \end{aligned}$