h) \( \operatorname{ctg} 15^{\circ}= \)

To find \( \operatorname{ctg} 15^{\circ} \), we can use the cotangent identity: \( \operatorname{ctg} x = \frac{1}{\tan x} \). We know \( \tan 15^{\circ} \) can be derived from the tangent subtraction formula: \[ \tan(15^{\circ}) = \tan(45^{\circ} - 30^{\circ}) = \frac{\tan 45^{\circ} - \tan 30^{\circ}}{1 + \tan 45^{\circ} \tan 30^{\circ}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} \] Simplifying, we get \( \tan(15^{\circ}) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \). Therefore: \[ \operatorname{ctg} 15^{\circ} = \frac{1}{\tan 15^{\circ}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \] After rationalizing this expression, you will find \( \operatorname{ctg} 15^{\circ} = 2 + \sqrt{3} \). There you go! Now you can impress your friends with your knowledge of trigonometric identities and angles!