Question

If $ABCD$ is a cyclic quadrilateral such that

$12\tan A-5=0$ and $5\cos B+3=0$, then $39(\cos C+tanD)=$

• A. $-16$
• B. $16$
• C. $-14$
• D. $14$

Answer 1

The correct option is B $16$

$12tanA-5=0$

$\therefore tanA=\frac{5}{12}$

$5cosB+3=0$

$\therefore cosB=\frac{-3}{5}$

$\angle A+\angle C={180}^{\circ }$

$\therefore \angle A={180}^{\circ }-\angle C$

$\therefore tanC=-tanA$

$\therefore tanC=\frac{-5}{12}$

$\therefore cosC=\frac{-12}{13}$ $({13}^2={12}^2+5^2)$

Now $\angle B+\angle D={180}^{\circ }$

$\therefore \angle B={180}^{\circ }-\angle D$

$\therefore cos\angle B=cos(180-\angle D)$

$\therefore cos\angle D=-cos\angle B$

$\therefore cosD=\frac{3}{5}$

$\therefore tanD=\frac{4}{3}$

$\therefore 39(cosC+tanD)$

$=39(\frac{-12}{13}+\frac{4}{3})$

$=52-36$

$=16$