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# 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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