If in $\Delta A B C, \cos ^{2} A+\cos ^{2} B+\cos ^{2} C=1$, prove that the triangle is right-angled.
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.
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