If $$\sec \theta - \tan \theta = \frac{x}{y}, (0 < x < y)$$ and $$0^\circ < \theta < 90^\circ$$, then $$\sin \theta$$ is equal to:

$$\sec \theta - \tan \theta = \frac{x}{y}$$ ---(1)

$$\frac{1}{\sec \theta - \tan \theta} = \frac{y}{x}$$

$$\frac{\sec^2 \theta - \tan^2 \theta}{\sec \theta - \tan \theta} = \frac{y}{x}$$
$$\frac{(\sec \theta - \tan \theta)(\sec \theta + \tan \theta)}{\sec \theta - \tan \theta} = \frac{y}{x}$$

$$\sec \theta + \tan \theta = \frac{y}{x}$$ ---(2)

From eq(1) and (2),

$$\sec \theta - \tan \theta + \sec \theta + \tan \theta =  \frac{x}{y} + \frac{y}{x}$$

$$2\sec \theta = \frac{x^2 + y^2}{xy} $$

$$\sec \theta = \frac{x^2 + y^2}{2xy} $$

$$\cos \theta = \frac{2xy}{x^2 + y^2} $$

Base = 2xy

Hypotenuses = {x^2 + y^2}

By Pythagoras,

$$(p)^2 + (2xy)^2 = (x^2 + y^2)^2$$

$$(p)^2 = x^4 + y^4 + 2x^2y^2 -  4x^2y^2

$$p = x^2 - y^2 = y^2 - x^2

sin$$\theta = p/h = \frac{y^2 - x^2}{x^2 + y^2}$$