If in ΔABC,

Solution:

Let ABC be any triangle.

In $\triangle \mathrm{ABC}$

$\cos ^{2} A+\cos ^{2} B+\cos ^{2} C=1$

$\Rightarrow \cos ^{2} A+\cos ^{2} B+\cos ^{2}[\pi-(B+A)]=1 \quad(\because A+B+C=\pi)$

 

$\Rightarrow \cos ^{2} A+\cos ^{2} B+\cos ^{2}(B+A)=1$

$\Rightarrow \cos ^{2} A+\cos ^{2} B=1-\cos ^{2}(B+A)$

 

$\Rightarrow \cos ^{2} A+\cos ^{2} B=\sin ^{2}(B+A)$

$\Rightarrow \cos ^{2} A+\cos ^{2} B=(\sin A \cos B+\cos A \sin B)^{2}$

 

$\Rightarrow \cos ^{2} A+\cos ^{2} B=\sin ^{2} A \cos ^{2} B+\cos ^{2} A \sin ^{2} B+2 \sin A \sin B \cos A \cos B$

$\Rightarrow \cos ^{2} A\left(1-\sin ^{2} B\right)+\cos ^{2} B\left(1-\sin ^{2} A\right)=2 \sin A \sin B \cos A \cos B$

 

$\Rightarrow 2 \cos ^{2} A \cos ^{2} B=2 \sin A \sin B \cos A \cos B$

$\Rightarrow \cos A \cos B=\sin A \sin B$

 

$\Rightarrow \cos A \cos B-\sin A \sin B=0$

$\Rightarrow \cos (A+B)=0$

 

$\Rightarrow \cos (A+B)=\cos 90^{\circ}$

$\Rightarrow A+B=90^{\circ}$

$\Rightarrow C=90^{\circ} \quad\left(\because A+B+C=180^{\circ}\right)$

Hence, $\triangle A B C$ is right angled.