If cos53° = x/y, then sec53° + cot37° is equal to:

Answer: Option C

$$\begin{array}{rl}& \cot 53^{\circ }=\frac{x}{y}=\frac{B}{H}\\ & P^2=H^2-B^2\\ & P=\sqrt{y^2-x^2}\\ & \sec 53^{\circ }+\cot 37^{\circ }\\ & =\sec 53^{\circ }+\tan 53^{\circ }\\ & =\frac{H}{B}+\frac{P}{B}\\ & =\frac{H+P}{B}\\ & =\frac{y+\sqrt{y^2-x^2}}{x}\end{array}$$