In a triangle $A B C$ where $∠ C = 90^{o}$ then prove that ${cos}^{2} A + {cos}^{2} B = 1$ .

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Solution

Given $△ A B C$ is right angled at $C$

Then $∠ C = 90^{o}$

So, $∠ A + ∠ B = 90^{o} . . . . . (1)$

Now, ${cos}^{2} A + {cos}^{2} B$

$= {cos}^{2} A + {cos}^{2} (90^{o} - A)$ (using (1))

$= {cos}^{2} A + {sin}^{2} A$

$= 1$

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