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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

यदि $x = \sec \phi - \tan \phi$ और $y = \csc \phi + \cot \phi$ है,तो निम्नलिखित में से कौन सा सत्य है?

यदि $\cos \theta + \sin \theta = \sqrt{2} \cos \theta$ है,तो $\cos \theta - \sin \theta =$

यदि $\cos (A - B) = \frac{3}{5}$ और $\tan A \tan B = 2$ है,तो

$\frac{\cos ^{2} 45^{\circ}}{\sin ^{2} 60^{\circ}}+\frac{\cos ^{2} 60^{\circ}}{\sin ^{2} 45^{\circ}}-\frac{\tan ^{2} 30^{\circ}}{\cot ^{2} 45^{\circ}}-\frac{\sin ^{2} 30^{\circ}}{\cot ^{2} 30^{\circ}}$ का संख्यात्मक मान है

$\sin ^4 \frac{\pi}{8} + \sin ^4 \frac{3\pi}{8} + \sin ^4 \frac{5\pi}{8} + \sin ^4 \frac{7\pi}{8} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module