If sin A + cos A = sqrt{2},then cos^2 A = | vedclass

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(B) Given: $\sin A + \cos A = \sqrt{2}$.Squaring both sides:$(\sin A + \cos A)^2 = (\sqrt{2})^2$$\sin^2 A + \cos^2 A + 2 \sin A \cos A = 2$Since $\sin

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$\frac{1}{2}$

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(B) Given: $\sin A + \cos A = \sqrt{2}$.Squaring both sides:$(\sin A + \cos A)^2 = (\sqrt{2})^2$$\sin^2 A + \cos^2 A + 2 \sin A \cos A = 2$Since $\sin^2 A + \cos^2 A = 1$,we have:$1 + 2 \sin A \cos A = 2$$2 \sin A \cos A = 1$$\sin 2A = 1 = \sin 90^\circ$$2A = 90^\circ \implies A = 45^\circ$.Now,calculate $\cos^2 A$:$\cos^2 45^\circ = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$.

If $\sin A + \cos A = \sqrt{2}$,then $\cos^2 A = $

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