In \( \Delta ABC \), right-angled at \( B \), if one angle is \( 45^\circ \), then the other angle is also \( 45^\circ \), i.e., \( \angle A = \angle C = 45^\circ \). Suppose \( BC = AB = a \). By Pythagoras Theorem, \( AC^2 = AB^2 + BC^2 = a^2 + a^2 = 2a^2 \), and therefore, \( AC = a\sqrt{2} \). Using the definitions of the trigonometric ratios, we have: \[ \sin 45^\circ = \frac{\text{side opposite to angle } 45^\circ}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}} \] \[ \cos 45^\circ = \frac{\text{side adjacent to angle } 45^\circ}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}} \] \[ \tan 45^\circ = \frac{\text{side opposite to angle } 45^\circ}{\text{side adjacent to angle } 45^\circ} = \frac{BC}{AB} = \frac{a}{a} = 1 \] Also, \( \csc 45^\circ = \frac{1}{\sin 45^\circ} = \sqrt{2} \), \( \sec 45^\circ = \frac{1}{\cos 45^\circ} = \sqrt{2} \), \( \cot 45^\circ = \frac{1}{\tan 45^\circ} = 1 \).