Correct Answer: $\frac{x^{2}}{y\sqrt{y^{2}–x^{2}}}$

Solution :
Given: $\sin17°=\frac{x}{y}$
BC = $\sqrt{y^{2}-x^{2}}$
So, $\sec17°=\frac{y}{\sqrt{y^{2}–x^{2}}}$ and $\sin73°=\frac{\sqrt{y^{2}–x^{2}}}{y}$
$\therefore(\sec17°–\sin73°)=\frac{y}{\sqrt{y^{2}–x^{2}}}–\frac{\sqrt{y^{2}–x^{2}}}{y}$= $\frac{y^{2}–(y^{2}–x^{2})}{y\sqrt{y^{2}–x^{2}}}$= $\frac{x^{2}}{y\sqrt{y^{2}–x^{2}}}$
Hence, the correct answer is $\frac{x^{2}}{y\sqrt{y^{2}–x^{2}}}$.