$\tan\alpha+\cot\alpha=2\\↔\cot\alpha=2-\tan\alpha$
Ta có: $\tan\alpha.\cot\alpha=1$
$→\tan\alpha.(2-\tan\alpha)=1\\↔-\tan^2\alpha+2\tan.\alpha=1\\↔\tan^2\alpha-2.\tan\alpha=-1\\↔\tan^2\alpha-2\tan\alpha+1=0\\↔(\tan\alpha-1)^2=0\\↔\tan\alpha=1\\↔\alpha=45^\circ$
Vậy $\alpha=45^\circ$