If A = sin 45^{circ} + cos 45^{circ} and B = | vedclass

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(A) We know that $A = \sin 45^{\circ} + \cos 45^{\circ} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.Alternatively,$A =

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$A > B$

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(A) We know that $A = \sin 45^{\circ} + \cos 45^{\circ} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.Alternatively,$A = \sqrt{2} (\frac{1}{\sqrt{2}} \sin 45^{\circ} + \frac{1}{\sqrt{2}} \cos 45^{\circ}) = \sqrt{2} \sin(45^{\circ} + 45^{\circ}) = \sqrt{2} \sin 90^{\circ} = \sqrt{2}(1) = \sqrt{2}$.Now,$B = \sin 44^{\circ} + \cos 44^{\circ} = \sqrt{2} (\frac{1}{\sqrt{2}} \sin 44^{\circ} + \frac{1}{\sqrt{2}} \cos 44^{\circ})$.Using the formula $\sin(x + y) = \sin x \cos y + \cos x \sin y$,we get $B = \sqrt{2} \sin(44^{\circ} + 45^{\circ}) = \sqrt{2} \sin 89^{\circ}$.Since $\sin 89^{\circ} Therefore,$B  B$.

If $A = \sin 45^{\circ} + \cos 45^{\circ}$ and $B = \sin 44^{\circ} + \cos 44^{\circ}$,then

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