In triangle ABC,if cos ^2 A + cos ^2 B + cos | vedclass

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Question

(C) Given $\cos ^2 A + \cos ^2 B + \cos ^2 C = 1$. Using the identity $\cos ^2 C = 1 - \sin ^2 C$,we have: $\cos ^2 A + \cos ^2 B + 1 - \sin ^2 C = 1$

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a right-angled triangle

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(C) Given $\cos ^2 A + \cos ^2 B + \cos ^2 C = 1$. Using the identity $\cos ^2 C = 1 - \sin ^2 C$,we have: $\cos ^2 A + \cos ^2 B + 1 - \sin ^2 C = 1$ $\Rightarrow \cos ^2 A + \cos ^2 B = \sin ^2 C$ Since $C = 180^{\circ} - (A + B)$,$\sin C = \sin(A + B)$. $\Rightarrow \cos ^2 A + \cos ^2 B = \sin ^2(A + B)$ $\Rightarrow \cos ^2 A + \cos ^2 B = (\sin A \cos B + \cos A \sin B)^2$ $\Rightarrow \cos ^2 A + \cos ^2 B = \sin ^2 A \cos ^2 B + \cos ^2 A \sin ^2 B + 2 \sin A \cos B \cos A \sin B$ $\Rightarrow \cos ^2 A(1 - \sin ^2 B) + \cos ^2 B(1 - \sin ^2 A) = 2 \sin A \cos B \cos A \sin B$ $\Rightarrow \cos ^2 A \cos ^2 B + \cos ^2 B \cos ^2 A = 2 \sin A \cos B \cos A \sin B$ $\Rightarrow 2 \cos ^2 A \cos ^2 B - 2 \sin A \cos B \cos A \sin B = 0$ $\Rightarrow 2 \cos A \cos B (\cos A \cos B - \sin A \sin B) = 0$ $\Rightarrow 2 \cos A \cos B \cos(A + B) = 0$ Since $\cos(A + B) = \cos(180^{\circ} - C) = -\cos C$,we get: $-2 \cos A \cos B \cos C = 0$ Thus,$\cos A = 0$ or $\cos B = 0$ or $\cos C = 0$. This implies $A = 90^{\circ}$ or $B = 90^{\circ}$ or $C = 90^{\circ}$. Therefore,$	riangle ABC$ is a right-angled triangle.

In $\triangle ABC$,if $\cos ^2 A + \cos ^2 B + \cos ^2 C = 1$,then $\triangle ABC$ is

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