If $\tan A = \cot B$,prove that $A + B = 90^{\circ}$.
(N/A) Given that,$\tan A = \cot B$.
We know that $\cot B = \tan(90^{\circ} - B)$.
Substituting this into the given equation,we get $\tan A = \tan(90^{\circ} - B)$.
Since the tangent functions are equal,their angles must be equal: $A = 90^{\circ} - B$.
Rearranging the terms,we get $A + B = 90^{\circ}$.
We know that $\cot B = \tan(90^{\circ} - B)$.
Substituting this into the given equation,we get $\tan A = \tan(90^{\circ} - B)$.
Since the tangent functions are equal,their angles must be equal: $A = 90^{\circ} - B$.
Rearranging the terms,we get $A + B = 90^{\circ}$.
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