If $\angle A$ and $\angle B$ are acute angles such that $\cos A = \cos B,$ then show that $\angle A = \angle B$.

(N/A) Let us consider a triangle $ABC$ in which $CD \perp AB$.
It is given that,
$\cos A = \cos B$
$\Rightarrow \frac{AD}{AC} = \frac{BD}{BC}$
$\Rightarrow \frac{AD}{BD} = \frac{AC}{BC}$
Let $\frac{AD}{BD} = \frac{AC}{BC} = k$
$\Rightarrow AD = k BD \dots(1)$
And,$AC = k BC \dots(2)$
Using Pythagoras theorem for triangles $CAD$ and $CBD,$ we obtain
$CD^2 = AC^2 - AD^2 \dots(3)$
And,$CD^2 = BC^2 - BD^2 \dots(4)$
From equations $(3)$ and $(4),$ we obtain
$AC^2 - AD^2 = BC^2 - BD^2$
$\Rightarrow (k BC)^2 - (k BD)^2 = BC^2 - BD^2$
$\Rightarrow k^2(BC^2 - BD^2) = BC^2 - BD^2$
$\Rightarrow k^2 = 1$
$\Rightarrow k = 1$
Putting this value in equation $(2),$ we obtain
$AC = BC$
$\Rightarrow \angle A = \angle B$ (Angles opposite to equal sides of a triangle are equal).