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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Updated on: 10-Oct-2022

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