From the following figure, find the value of:
${cos}^{2} A + {sin}^{2} A$

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Solution

Given, $△ A B C$ , $∠ A B C = 90^{\circ}$ , $A B = B C = a$
By Pythagoras Theorem,
$A C^{2} = A B^{2} + B C^{2}$
$A C^{2} = a^{2} + a^{2}$
$A C = a \sqrt{2}$
Now, ${cos}^{2} A + {sin}^{2} A = (\frac{B}{H})^{2} + (\frac{P}{H})^{2} = (\frac{a}{a \sqrt{2}})^{2} + (\frac{a}{a \sqrt{2}})^{2}$
= $\frac{1}{2} + \frac{1}{2}$
= $1$

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