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# If $$tan A \ = \ cot B$$ , prove that $$A + B \ = \ 90°$$.

Given, $$tan A \ = \ cot B$$
We know that, $$cot \theta \ = \ tan(90° – \theta)$$

=> $$cot B \ = \ tan(90° - B)$$

$$tan A \ = \ tan(90° - B)$$

=> $$(90° – A) \ = \ B$$ [$$\because$$ (90° – A) and B are both acute angles ]

=> $$A + B \ = \ 90°$$