If $tan A = cot B$ , prove that $A + B + 90$ , where A and B are measures of acute angles.

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Solution

According to question,

$tan A = cot B$

$⟹ tan A = tan (90 - B)$

As, $A$ , and $B$ , are acute angles,

Then,

$⟹ A = 90 - B$

$⟹ A + B = 90$

Hence proved.

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Similar questions

∠ A

∠ B

∠ A = ∠ B

A = cot B

A

B

A + B = 90^{\circ} .

\sqrt 3

\frac{1}{\sqrt{3}} .

\tan A = \frac{1}{2}, \tan B = \frac{1}{3} and \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}

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