Question

If $ABCD$ (in order) is a quadrilateral inscribed in a circle, then which of the following is/are always true?

• A. $\sec B=\sec D$
• B. $\cos A+\cos C=0$
• C. $\text{cosec}A=\text{cosec}C$
• D. $\tan B+\tan D=0$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $\cos A+\cos C=0$
C $\text{cosec}A=\text{cosec}C$
D $\tan B+\tan D=0$

Opposite angles of a cyclic quadrilateral are supplementry
So, $A+C=\pi$ and $B+D=\pi$
$\Rightarrow A=\pi -C$ and $B=\pi -D$

Now from the options

Option $\left(a\right)$ From $B=\pi -D\Rightarrow \sec B=\sec (\pi -D)$
$\Rightarrow \sec B=-\sec D$

Option $\left(b\right)$ From $A=\pi -C\Rightarrow \cos A=\cos (\pi -C)$
$\Rightarrow \cos A+\cos C=0$

Option $\left(c\right)$ From $A=\pi -C\Rightarrow \text{cosec}A=\text{cosec}(\pi -C)$
$\Rightarrow \text{cosec}A=\text{cosec}C$

Option $\left(d\right)$ From $B=\pi -D\Rightarrow \tan B=\tan (\pi -D)$
$\Rightarrow \tan B=-\tan D$