If $A B C D$ is a cyclic quadrilateral such that $12$ $tan A - 5 = 0$ and 5 $cos B + 3 = 0$ , then $cos C tan D =$

\frac{- 16}{13}

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^{\frac{16}{13}}

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\frac{- 13}{16}

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\frac{23}{16}

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Solution

The correct option is A $\frac{- 16}{13}$
$cos A$ $= \frac{12}{13}$
and $tan B$ $= - \frac{4}{3}$
Sum of opposite angles of a cyclic quadrilateral is $180^{\circ}$
So $C$ = $180$ $^{\circ}$ - $A$ and $D$ = $180$ $^{\circ} - B$
$cos C tan D$ $= cos (180^{\circ} - A) tan (180^{\circ} - B)$
$= (- cos A) (- tan B) = cos A tan B$ = $\frac{- 16}{13}$

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