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

Let us consider two right triangles ACD and BEF where cos A = cos B.

We know that,

cos A = $\dfrac{\text{Side adjacent to angle A}}{\text{Hypotenuse}} = \dfrac{AC}{AD}$

cos B = $\dfrac{\text{Side adjacent to angle B}}{\text{Hypotenuse}} = \dfrac{BE}{BF}$.

Given,

⇒ cos A = cos B

⇒ $\dfrac{AC}{AD} = \dfrac{BE}{BF}$

⇒ $\dfrac{AC}{BE} = \dfrac{AD}{BF}$ = k (let) ……..(1)

⇒ AC = k.BE and AD = k.BF

In right angle triangle ACD,

By pythagoras theorem,

⇒ AD² = AC² + CD²

⇒ CD² = AD² - AC²

⇒ CD² = (k.BF)² - (k.BE)²

⇒ CD² = k²(BF² - BE²)

⇒ CD = $\sqrt{k^2(BF^2 - BE^2)} = k\sqrt{(BF^2 - BE^2)}$

In right angle triangle BEF,

By pythagoras theorem,

⇒ BF² = BE² + EF²

⇒ EF² = BF² - BE²

⇒ EF = $\sqrt{BF^2 - BE^2}$

So,

$\dfrac{CD}{EF} = \dfrac{k\sqrt{BF^2 - BE^2}}{\sqrt{BF^2 - BE^2}}$ = k ……….(2)

From (1) and (2), we get :

$\dfrac{AC}{BE} = \dfrac{AD}{BF} = \dfrac{CD}{FE}$

Since, ratio of corresponding sides of similar triangle are proportional.

∴ △ACD ~ △BEF.

∴ ∠A = ∠B.

Hence, proved that ∠A = ∠B.