Given: \(\cos 43^\circ = \frac{x}{\sqrt{x^2 + y^2}}\) Then, \[ \sec 43^\circ = \frac{\sqrt{x^2 + y^2}}{x} \] Since \(\sec(90^\circ - 47^\circ) = \sec 43^\circ\), \[ \cosec 47^\circ = \frac{\sqrt{x^2 + y^2}}{x} \] Squaring both sides: \[ \cosec^2 47^\circ = \left( \frac{\sqrt{x^2 + y^2}}{x} \right)^2 = \frac{x^2 + y^2}{x^2} \] Now, \[ \cosec^2 47^\circ - 1 = \frac{x^2 + y^2}{x^2} - 1 = \frac{y^2}{x^2} \] So, \[ \cot^2 47^\circ = \frac{y^2}{x^2} \Rightarrow \tan^2 47^\circ = \frac{x^2}{y^2} \Rightarrow \tan 47^\circ = \frac{x}{y} \] (Answer)