$16 cos 144^{0} . cos 108^{0} . cos 72^{0} . cos 36^{0} =$

5

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Solution

The correct option is C $1$
$16 cos 36^{0} cos 108^{0} cos 72^{0} cos 144^{0}$

$= \frac{8}{sin 36^{0}} sin 72^{0} cos 72^{0} cos 144^{0} cos 108^{0}$

$= \frac{4}{sin 36^{0}} sin 144^{0} cos 108^{0} cos 144^{0}$

$= \frac{2 sin 288^{0} cos 108^{0}}{sin 36^{0}}$

$= \frac{- 2 sin 72^{0} cos 108^{0}}{sin 36^{0}}$

$= \frac{- sin 180^{0} - sin (- 36^{0})}{sin 36^{0}}$

$= 1$

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