Question

If ABCD is a cyclic quadrilateral then $\tan A+\tan C$ is

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

Answer: (A)

$0$

Given that : $ABCD$ is a cyclic quadrilateral.

According to Cyclic Quadrilateral property, 'Sum of the opposite angles of a cyclic quadrilateral is ${180}^{\circ }$'.

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

$\therefore$ $\angle$$A={180}^{\circ }-$$\angle$$C$ ....$\left(i\right)$

According to question we need to find the value of $(tanA+tanC)$

$tanA+tanC$

$=tan(180-C)+tanC$ ...From $eq.\left(i\right)$

[Since we know that, $tan({180}^{\circ }-\theta )=-tan\theta$

So, $tan(180-C)=-tanC$]

$\therefore$ $tan(180-C)+tanC$

$=-tanC+tanC$

$=0$

Hence, option $A$ is correct.