Question

If ABCD is a cyclic quadrilateral,$\tan B-\tan D=2\sqrt{3}$ then $\tan 3B=$

• A. $0$
• B. $-1$
• C. $-2$
• D. $2$

Answer 1

The correct option is A $0$

$ABCD$ is a cyclic quadrilateral.

$\therefore$ $\angle B+\angle D={180}^o$ [ Sum of opposite angles of cyclic quadrilateral is supplementary ]

$\Rightarrow$ $\angle D={180}^o-\angle B$

$\Rightarrow$ $D={180}^o-B$ ----- ( 1 )

$\Rightarrow$ $\tan B-\tan D=2\sqrt{3}$ [ Given ]

$\Rightarrow$ $\tan B-\tan ({180}^o-B)=2\sqrt{3}$ [ From ( 1 ) ]

$\Rightarrow$ $\tan B+\tan B=2\sqrt{3}$ [ Since $\tan ({180}^o-\theta )=-\tan \theta$ ]

$\Rightarrow$ $2\tan B=2\sqrt{3}$

$\Rightarrow$ $\tan B=\sqrt{3}$

$\therefore$ $B={\tan }^{-1}\left(\sqrt{3}\right)$

$\Rightarrow$ $B={60}^o$

$\Rightarrow$ $\tan 3B=\tan (3\times {60}^o)$

$\Rightarrow$ $\tan 3B=\tan \left({180}^o\right)$

$\Rightarrow$ $\tan 3B=0$