Question

If $A=\sin {45}^{\circ }+\cos {45}^{\circ }$ and $B=\sin {44}^{\circ }+\cos {44}^{\circ }$. Which of the following holds true?

• A. $A>B$
• B. $A<B$
• C. $A=B$
• D. none of these

Answer 1

The correct option is A $A>B$

$A=\sin {45}^{\circ }+\cos {45}^{\circ }\&B=\sin {44}^{\circ }+\cos {44}^{\circ }$
$A-B=(\sin {45}^{\circ }-\sin {44}^{\circ })+(\cos {45}^{\circ }-\cos {44}^{\circ })\Rightarrow A-B=2\cos \left(\frac{{45}^{\circ }{+44}^{\circ }}{2}\right)\sin \left(\frac{{45}^{\circ }{-44}^{\circ }}{2}\right)-2\sin \left(\frac{{45}^{\circ }{+44}^{\circ }}{2}\right)\sin \left(\frac{{45}^{\circ }{-44}^{\circ }}{2}\right)\Rightarrow A-B=2\sin \frac{1^{\circ }}{2}(\cos \frac{{99}^{\circ }}{2}-\sin \frac{{99}^{\circ }}{2})$
For $0<\theta <\frac{\pi }{4}$

$\cos \theta >\sin \theta$
$\therefore A-B>0\Rightarrow A>B$
Ans: A