If A=sin⁡45∘+cos⁡45∘ and B=sin⁡44∘+cos⁡44∘,then

A
A > B
B
A < B
C
A = B
D
A = 2B

Solution:

$\begin{array}{l} A = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \sin 90^{0} \\ \Rightarrow B = \sqrt{2} [\frac{1}{\sqrt{2}} \sin 44^{0} + \frac{1}{\sqrt{2}} \cos 44^{0}] \\ = \sqrt{2} \sin (45^{0} + 44^{0}) \\ = \sqrt{2} \sin 89^{0} < \sqrt{2} \sin 90^{\circ} ∴ A > B \end{array}$

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