Question

If $ABCD$ is a cyclic quadrilateral, then the value of $\tan A+\tan B+\tan C+\tan D$ is
$(A,B,C,D\ne \frac{\pi }{2})$

• A. $-1$
• B. $1$
• C. $0$
• D. None of these

Answer 1

The correct option is C $0$

$\because ABCD$ is cyclic quadrilateral
$\therefore A+C=\pi$ and $B+D=\pi$
$\Rightarrow A=\pi -C$ and $B=\pi -D$

Now,
$\tan A+\tan B+\tan C+\tan D=\tan (\pi -C)+\tan (\pi -D)+\tan C+\tan D=-\tan C-\tan D+\tan C+\tan D=0$