Question

ABCD a cyclic quadrilateral such that $12tanA-5=0$ and $5cosB+3=0$, Then $39(cosC+tanD)=$

• A. -16
• B. 16
• C. -14
• D. 14

Answer 1

The correct option is A -16

ABCD is a cyclic quadrilateral

$\angle A+\angle C=180$ degree and $\angle B+\angle D=180$ degree

As of these no angle of a cyclic quadrilateral can be >180 degree.

(ii) Given: $12\times tan\left(A\right)=0$; So $ tan (A) = 5/12 $

As $A+C=180$ degree

$C=180-A$

$tan\left(C\right)=tan(180-A)=-tan\left(A\right)=-5/12$

so $cos\left(C\right)=-12/13$

(iii) $cos\left(B\right)=-3/15$;

$cos\left(D\right)=cos(180-B)=-cos\left(B\right)=3/5$

$tan\left(D\right)=4/3$

(iv) Sum of roots $=cos\left(C\right)+tan\left(D\right)=-12/13+4/3=-16/39$

then $39(cosC+tanD)=-16$

So, the correct option is $-16$.