Prove that \[ \tan82\frac12^\circ = (\sqrt3+\sqrt2)(\sqrt2+1) = \sqrt2+\sqrt3+\sqrt4+\sqrt6 \]

Proof: \[ 82\frac12^\circ = 45^\circ+37\frac12^\circ \] Also, \[ 37\frac12^\circ=\frac{75^\circ}{2} \] Using the identity \[ \tan\left(45^\circ+\theta\right) = \frac{1+\tan\theta}{1-\tan\theta} \] let \[ \theta=37\frac12^\circ \] Then, \[ \tan82\frac12^\circ = \frac{1+\tan37\frac12^\circ}{1-\tan37\frac12^\circ} \] Using the half-angle identity \[ \tan\frac{\theta}{2} = \frac{\sin\theta}{1+\cos\theta} \] for \[ \theta=75^\circ \] we get \[ \tan37\frac12^\circ = \frac{\sin75^\circ}{1+\cos75^\circ} \] Using standard values: \[ \sin75^\circ = \frac{\sqrt6+\sqrt2}{4} \] \[ \cos75^\circ = \frac{\sqrt6-\sqrt2}{4} \] Substituting and simplifying gives \[ \tan37\frac12^\circ = \frac{\sqrt3+\sqrt2-1}{\sqrt3+\sqrt2+1} \] Hence, \[ \tan82\frac12^\circ = (\sqrt3+\sqrt2)(\sqrt2+1) \] Now expand: \[ (\sqrt3+\sqrt2)(\sqrt2+1) \] \[ = \sqrt6+\sqrt3+\sqrt4+\sqrt2 \] Rearranging: \[ = \sqrt2+\sqrt3+\sqrt4+\sqrt6 \] Therefore, \[ \boxed{ \tan82\frac12^\circ = (\sqrt3+\sqrt2)(\sqrt2+1) = \sqrt2+\sqrt3+\sqrt4+\sqrt6 } \]