$\frac{\sin 70 ° + \cos 40 °}{\cos 70 ° + \sin 40 °} =$

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$- \sqrt{3}$

$\sqrt{3}$

$- \sqrt{2}$

$\sqrt{2}$

answer is B.

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Detailed Solution

$\frac{\sin 70 ° + \cos 40 °}{\cos 70 ° + \sin 40 °}$

$= \frac{\sin 70 ° + \cos (90 ° - 50 °)}{\cos 70 ° + \sin (90 ° - 50 °)}$

$= \frac{\sin 70 ° + \sin 50 °}{\cos 70 ° + \cos 50 °}$

$= \frac{2 \sin (\frac{70 ° + 50 °}{2}) \cos (\frac{70 ° - 50 °}{2})}{2 \cos (\frac{70 ° + 50 °}{2}) \cos (\frac{70 ° - 50 °}{2})}$

$= \frac{\sin 60 °}{\cos 60 °}$

$= \tan 60 °$

$= \sqrt{3}$

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