Use a sum or difference formula to find the exact value of the trigonometric function. \( \tan 105^{\circ} \) \( \tan 105^{\circ}=\square \) (Type an exact answer, using radicals as needed. Simplify your answer.)

To find \( \tan 105^{\circ} \), we can use the angle sum formula for tangent. We can express \( 105^{\circ} \) as \( 60^{\circ} + 45^{\circ} \): \[ \tan(105^{\circ}) = \tan(60^{\circ} + 45^{\circ}) = \frac{\tan(60^{\circ}) + \tan(45^{\circ})}{1 - \tan(60^{\circ})\tan(45^{\circ})} \] Now we know that \( \tan(60^{\circ}) = \sqrt{3} \) and \( \tan(45^{\circ}) = 1 \): \[ = \frac{\sqrt{3} + 1}{1 - \sqrt{3}\cdot1} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} \] To simplify this expression, multiply the numerator and the denominator by the conjugate of the denominator \( 1 + \sqrt{3} \): \[ = \frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} \] \[ = \frac{(\sqrt{3} + 1 + 3 + \sqrt{3})}{1 - 3} = \frac{4 + 2\sqrt{3}}{-2} = -2 - \sqrt{3} \] Thus, the exact value of \( \tan 105^{\circ} \) is: \[ \tan 105^{\circ} = -2 - \sqrt{3} \]