If sin21° = x/y , then sec21° - sin69° is equal to?

Answer: Option A

$$\begin{array}{rl}& \text{In }△\text{ABC sin2}{\text{1}}^{\circ }\text{ = }\frac{x}{y}\\ & \text{AB}=x\\ & \text{AC}=y\\ & \text{BC}=\sqrt{y^2-x^2}\\ & \Rightarrow \text{sec2}{\text{1}}^{\circ }-\sin 69^{\circ }\\ & \Rightarrow \frac{\text{AC}}{\text{BC}}-\frac{\text{BC}}{\text{AC}}\\ & \Rightarrow \frac{{\left(\text{AC}\right)}^2-{\left(\text{BC}\right)}^2}{\left(\text{BC}\right)\left(\text{AC}\right)}=\frac{y^2-{\left(\sqrt{\left(y^2-x^2\right)}\right)}^2}{y\sqrt{y^2-x^2}}\\ & \Rightarrow \frac{y^2-y^2+x^2}{y\sqrt{y^2+x^2}}=\frac{x^2}{y\sqrt{y^2-x^2}}\end{array}$$