If $tan\ A = cot\ B$, prove that $A + B = 90^o$.
Given:
$tan\ A=cot\ B$
To do:
We have to prove that $A+B=90^o$.
Solution:
We know that,
$cot\ B=tan\ (90^o-B)$ Therefore,
$tan\ A=tan\ (90^o-B)$
This implies,
$A=90^o-B$
$A+B=90^o$
Hence proved.
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