If $\tan A =\cot B ,$ prove that $A + B =90^{\circ}$
If $\tan A =\cot B ,$ prove that $A + B =90^{\circ}$
Given that,
$\tan A =\cot B$
$\tan A=\tan \left(90^{\circ}-B\right)$
$A=90^{\circ}-B$
$A+B=90^{\circ}$
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