If $sin17^{0}$ = $\frac{x}{y}$, then $sec 17^{0} - sin 73^{0}$ is equal to
1). $\frac{y}{(\sqrt{y^{2}-x^{2}})}$
2). $\frac{y^{2}}{(x\sqrt{y^{2}-x^{2}})}$
3). $\frac{x}{(y\sqrt{y^{2}-x^{2}})}$
4). $\frac{x^{2}}{(y\sqrt{y^{2}-x^{2}})}$
sin 17° = $\frac{x}{y}$
sin 73° = sin (90° – 17°)= cos 17°
$\therefore$ cos 17° = $\sqrt{1-sin^{2}17°}$ = $\sqrt{1-\frac{x^{2}}{y^{2}}}$
= $\frac{\sqrt{y^{2}-x^{2}}}{y}$ .