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Question
If in \[∆ ABC, \cos^2 A + \cos^2 B + \cos^2 C = 1\] prove that the triangle is right-angled.
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Solution
Let ABC be any triangle.
In \[∆ ABC\]
\[\cos^2 A + \cos^2 B + \cos^2 C = 1\]
\[ \Rightarrow \cos^2 A + \cos^2 B + \cos^2 \left[ \pi - \left( B + A \right) \right] = 1 \left( \because A + B + C = \pi \right)\]
\[ \Rightarrow \cos^2 A + \cos^2 B + \cos^2 \left( B + A \right) = 1\]
\[ \Rightarrow \cos^2 A + \cos^2 B = 1 - \cos^2 \left( B + A \right)\]
\[ \Rightarrow \cos^2 A + \cos^2 B = \sin^2 \left( B + A \right)\]
\[ \Rightarrow \cos^2 A + \cos^2 B = \left( \sin A\cos B + \cos A\sin B \right)^2 \]
\[ \Rightarrow \cos^2 A + \cos^2 B = \sin^2 A \cos^2 B + \cos^2 A \sin^2 B + 2\sin A\sin B\cos A\cos B\]
\[ \Rightarrow \cos^2 A\left( 1 - \sin^2 B \right) + \cos^2 B\left( 1 - \sin^2 A \right) = 2\sin A\sin B\cos A\cos B\]
\[ \Rightarrow 2 \cos^2 A \cos^2 B = 2\sin A\sin B\cos A\cos B\]
\[ \Rightarrow \cos A\cos B = \sin A\sin B\]
\[ \Rightarrow \cos A\cos B - \sin A\sin B = 0\]
\[ \Rightarrow \cos \left( A + B \right) = 0\]
\[ \Rightarrow \cos \left( A + B \right) = \cos {90}^° \]
\[ \Rightarrow A + B = {90}^°\]
\[ \Rightarrow C = {90}^° \left( \because A + B + C = {180}^° \right)\]
Hence, \[∆\]ABC is right angled.
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