Catcolate the value of \( \sin 50^{\circ}+\cos 50^{\circ} \tan 20^{\circ} \) without the use of a calculator.

To solve for \( \sin 50^{\circ}+\cos 50^{\circ} \tan 20^{\circ} \), we start by noting the trigonometric identity for tangent: \[ \tan 20^{\circ} = \frac{\sin 20^{\circ}}{\cos 20^{\circ}}. \] Substituting this into our expression gives: \[ \sin 50^{\circ} + \cos 50^{\circ} \cdot \frac{\sin 20^{\circ}}{\cos 20^{\circ}} = \sin 50^{\circ} + \frac{\cos 50^{\circ} \sin 20^{\circ}}{\cos 20^{\circ}}. \] Next, we use the co-function identity \( \sin(90^\circ - x) = \cos x \) to find that \( \sin 50^{\circ} = \cos 40^{\circ} \). Thus, the expression becomes: \[ \cos 40^{\circ} + \frac{\cos 50^{\circ} \sin 20^{\circ}}{\cos 20^{\circ}}. \] However, it is more convenient to manipulate the components: Noting that \( \cos 50^{\circ} = \sin 40^{\circ} \), we can also rewrite the second term: \[ \sin 40^{\circ} \cdot \tan 20^{\circ} = \sin 40^{\circ} \cdot \frac{\sin 20^{\circ}}{\cos 20^{\circ}}. \] Based on the identity \( \sin a \cdot \tan b = \frac{\sin a \cdot \sin b}{\cos b} \), we write: \[ \sin 40^{\circ} \cdot \tan 20^{\circ} = \frac{\sin 40^{\circ} \cdot \sin 20^{\circ}}{\cos 20^{\circ}}. \] For \( \sin 40^{\circ} = 2 \sin 20^{\circ} \cos 20^{\circ} \), this simplifies our function further: Combining these gives: \[ \sin 50^{\circ} + \cos 50^{\circ} \tan 20^{\circ} = \sin 50^{\circ} + \sin 40^{\circ} = \sin 50^{\circ} + 2\sin 20^{\circ} \cos 20^{\circ} / \cos 20^{\circ} = 1. \] Thus: \[ \sin 50^{\circ} + \cos 50^{\circ} \tan 20^{\circ} = 1. \] Therefore, the answer is \( \boxed{1} \).