Question

If $tanA=\cot B,$ prove that $A+B={90}^0$

Answer 1

$\tan A=\cos B$ $\cot \theta =\tan (90-\theta )$

$\Rightarrow \tan A=\tan (90-B)$

$\Rightarrow \tan A=\tan (90-B)$

on solving

$\Rightarrow A=90-B$

$\Rightarrow A+B=90$

Hence, proved.