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

$\frac{y^2-x^2}{x^2+y^2}$

secθ - tanθ = \(\frac{x}{y}\)      -----(1)

{ using , sec²θ - tan²θ = 1 , So, secθ + tanθ = \(\frac{1}{secθ - tanθ}\) }

secθ + tanθ = \(\frac{y}{x}\)        -----(2)

On adding equation 1 and 2.

2 secθ = \(\frac{x}{y}\) + \(\frac{y}{x}\) 

2 secθ = \(\frac{x² + y² }{xy}\)

secθ = \(\frac{x² + y² }{ 2xy}\)

cosθ = \(\frac{2xy }{ x² + y²}\)

{ using , sin²θ + cos²θ = 1 }

sin²θ = 1 - cos²θ

= 1 -  \(\frac{2xy }{ x² + y²}\)

= 1 - \(\frac{4x²y² }{ (x² + y²)² }\)

= \(\frac{ (x² + y²)² - 4x²y² }{ (x² + y²)² }\)

= \(\frac{ ( y² - x²)² }{ (x² + y²)² }\)

sinθ = \(\frac{ ( y² - x²) }{ (x² + y²) }\)