Question

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

Answer 1

Let ABC be any triangle.
In $∆\mathrm{ABC}$,
$$\begin{gathered} {\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C=1 \\ \Rightarrow {\cos }^{2}A+{\cos }^{2}B+{\cos }^2\left[\mathrm{\pi }-\left(B\mathit{+}A\right)\right]=1\left(\because A+B+C=\mathrm{\pi }\right) \\ \Rightarrow {\cos }^{2}A+{\cos }^{2}B+{\cos }^2\left(B\mathit{+}A\right)=1 \\ \Rightarrow {\cos }^{2}A+{\cos }^{2}B=1-{\cos }^2\left(B\mathit{+}A\right) \\ \Rightarrow {\cos }^{2}A+{\cos }^{2}B={\sin }^2\left(B\mathit{+}A\right) \\ \Rightarrow {\cos }^{2}A+{\cos }^{2}B={\left(\sin A\cos B+\cos A\sin B\right)}^2 \\ \Rightarrow {\cos }^{2}A+{\cos }^{2}B={\sin }^{2}A{\cos }^{2}B+{\cos }^{2}A{\sin }^{2}B+2\sin A\sin B\cos A\cos B \\ \Rightarrow {\cos }^{2}A\left(1-{\sin }^{2}B\right)+{\cos }^{2}B\left(1-{\sin }^{2}A\right)=2\sin A\sin B\cos A\cos B \\ \Rightarrow 2{\cos }^{2}A{\cos }^{2}B=2\sin A\sin B\cos A\cos B \\ \Rightarrow \cos A\cos B=\sin A\sin B \\ \Rightarrow \cos A\cos B-\sin A\sin B=0 \\ \Rightarrow \cos \left(A+B\right)=0 \\ \Rightarrow \cos \left(A+B\right)=\cos {90}^{°} \\ \Rightarrow A+B={90}^{°} \\ \Rightarrow C={90}^{°}\left(\because A+B+C={180}^{°}\right) \end{gathered}$$

Hence, $∆$ABC is right angled.