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Maths-

General

Easy

Question

If ABCD is a cynic quadrilateral such that 12 tan $A minus 5 equals 0$ and 5cos B+3=0, then cos C tan D=

$\frac{- 16}{13}$
$\frac{16}{13}$
$\frac{- 13}{16}$
$\frac{23}{16}$

hint Hint:

For ABCD to be a cynic quadrilateral A+C = $straight pi$ and B+D = $straight pi$ .

The correct answer is: $fraction numerator negative 16 over denominator 13 end fraction$

Step by step solution:
Given, ABCD is a cynic quadrilateral
$12 space tan A minus 5 space equals 0$
$rightwards double arrow tan A equals 5 over 12$
$5 space cos B plus 3 space equals 0$
$rightwards double arrow cos B equals negative 3 over 5$
Now, A+C = $straight pi$
Therefore, C = $straight pi$ -A
Thus, $cos C space equals space cos left parenthesis straight pi$ $negative straight A right parenthesis space equals space minus space cosA$ $equals negative 12 over 13$
Similarly, B+D = $straight pi$
Therefore, D= $straight pi$ -B
Thus, $tan D equals tan left parenthesis straight pi$ $negative straight B right parenthesis equals space minus space tanB$ $equals negative 4 over 3$
Therefore, $cos C space tan B$ $equals negative 12 over 13 cross times negative 4 over 3$ $equals 16 over 13$
Hence, option(b) is the correct option.

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