If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Let we have a triangle, 

As per the given condition, A and B are acute angles, such that

cos (A) = cos (B)

Since, $Cos \theta =\frac{Adjacent side}{Hypotenuse}$

So, $\frac{AC}{AB}= \frac{BC}{AB}$

Solving further, we get, 

$AC = BC$

$\angle A=\angle B$ (Angles opposite to equal side of an isosceles triangle are equal)