यदि A = sin 45^{circ} + cos 45^{circ} और B = | vedclass

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(A) हम जानते हैं कि $A = \sin 45^{\circ} + \cos 45^{\circ} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.वैकल्पिक रूप से,

Vedclass · Accepted Answer

$A > B$

Vedclass · Answer

(A) हम जानते हैं कि $A = \sin 45^{\circ} + \cos 45^{\circ} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.वैकल्पिक रूप से,$A = \sqrt{2} (\frac{1}{\sqrt{2}} \sin 45^{\circ} + \frac{1}{\sqrt{2}} \cos 45^{\circ}) = \sqrt{2} \sin(45^{\circ} + 45^{\circ}) = \sqrt{2} \sin 90^{\circ} = \sqrt{2}(1) = \sqrt{2}$.अब,$B = \sin 44^{\circ} + \cos 44^{\circ} = \sqrt{2} (\frac{1}{\sqrt{2}} \sin 44^{\circ} + \frac{1}{\sqrt{2}} \cos 44^{\circ})$.सूत्र $\sin(x + y) = \sin x \cos y + \cos x \sin y$ का उपयोग करने पर,हमें $B = \sqrt{2} \sin(44^{\circ} + 45^{\circ}) = \sqrt{2} \sin 89^{\circ}$ प्राप्त होता है।चूंकि $\sin 89^{\circ} अतः,$B  B$।

यदि $A = \sin 45^{\circ} + \cos 45^{\circ}$ और $B = \sin 44^{\circ} + \cos 44^{\circ}$ है,तो

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