$\cos 70^{\circ} + \sin 40^{\circ} =$

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$\sin 10 °$

$\tan 10 °$

$\cos 10 °$

$c o t 10 °$

answer is B.

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Detailed Solution

$\cos 70^{\circ} + \sin 40^{\circ} = \cos 70^{\circ} + \cos 50 ° = 2 \cos \frac{70^{\circ} + 50 °}{2} \cos \frac{70^{\circ} - 50 °}{2} = 2 \cos 60 ° \cos 10 ° = \cos 10 °$

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