Question

Suppose $ABCD$ (in order) is a quadrilateral inscribed in a circle. Which of the following is/are always true?

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

Answer 1

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

$\tan B+\tan D=0$
$\text{cosec A}=\text{cosec C}$
$\cot A+\cot C=0$

From the figure:

$Arc\left(BCD\right)=2A$ and$Arc\left(DAB\right)=2C$

$Arc\left(BCD\right)+Arc\left(DAB\right)=$ $2\pi$

$\Rightarrow 2A+2C=2\pi$

Therefore $A+C=\pi$

Similarly $B+D=\pi$

$\because A=\pi -C\Rightarrow \cot A=\cot (\pi -C)\Rightarrow \cot A+\cot C=0$

and $\csc A=\csc (\pi -C)$

$\Rightarrow \csc A=\csc C$

$\sec B=\sec (\pi -D)$

$\Rightarrow \sec B=-\sec D$

and $\tan B=\tan (\pi -D)$

$\Rightarrow \tan B+\tan D=0$

Options B,C,D are correct.