CUET Preparation Today
If $\cos 53^{\circ}=\frac{x}{y}$, then $\sec 53^{\circ}+\cot 37^{\circ}$ is equal to: |
$\frac{x+\sqrt{y^2-x^2}}{y}$ $\frac{x+\sqrt{y^2-x^2}}{x}$ $\frac{y+\sqrt{y^2-x^2}}{x}$ $\frac{y+\sqrt{y^2-x^2}}{y}$ |
$\frac{y+\sqrt{y^2-x^2}}{x}$ |
