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

Let us consider a triangle $ABC$ in which $CD \perp AB$.

It is given that,

$\cos A=\cos B$

$\Rightarrow \frac{A D}{A C}=\frac{B D}{B C}$

$\Rightarrow \frac{A D}{B D}=\frac{A C}{B C}$

Let $\frac{A D}{B D}=\frac{A C}{B C}=k$

$\Rightarrow AD =k BD \ldots(1)$

And, $A C=k B C \ldots(2)$

Using Pythagoras theorem for triangles $CAD$ and $CBD,$ we obtain

$CD ^{2}= AC ^{2}- AD ^{2} \ldots(3)$

And, $CD ^{2}= BC ^{2}- BD ^{2} \ldots(4)$

From equations $( 3 )$ and $(4),$ we obtain

$AC ^{2}- AD ^{2}= BC ^{2}- BD ^{2}$

$\Rightarrow(k BC )^{2}-(k BD )^{2}= BC ^{2}- BD ^{2}$

$\Rightarrow k^{2}\left(B C^{2}-B D^{2}\right)=B C^{2}-B D^{2}$

$\Rightarrow k^{2}=1$

$\Rightarrow k=1$

Putting this value in equation $(2),$ we obtain

$AC = BC$

$\Rightarrow \angle A=\angle B$ (Angles opposite to equal sides of a triangle)