Question

Which of the following statements hold true for a quadrilateral ABCD?

• A. $sin\left(\frac{A+B}{4}\right)=cot\left(\frac{C+D}{4}\right)$
• B. $sin\left(\frac{A+B}{4}\right)=cos\left(\frac{C+D}{4}\right)$
• C. $tan\left(\frac{A+C}{4}\right)=cot\left(\frac{B+D}{4}\right)$
• D. $tan\left(\frac{A+C}{4}\right)=cos\left(\frac{B+D}{4}\right)$

Answer 1

Answer: (B), (C)

The correct options are
B $sin\left(\frac{A+B}{4}\right)=cos\left(\frac{C+D}{4}\right)$
C $tan\left(\frac{A+C}{4}\right)=cot\left(\frac{B+D}{4}\right)$

In a quadrilateral, sum of all the interior angles is 360${}^{\circ }$.
$A+B+C+D={360}^{\circ }$
$⟹\frac{A+B+C+D}{4}={90}^{\circ }$
$⟹\frac{A+B}{4}={90}^{\circ }-\frac{C+D}{4}$
$⟹sin\left(\frac{A+B}{4}\right)=sin({90}^{\circ }-\frac{C+D}{4})$
Since, $cos\theta =sin({90}^{\circ }-\theta )$
$⟹sin\left(\frac{A+B}{4}\right)=cos\left(\frac{C+D}{4}\right)$
Now, let's simplify this in a different way.
$\frac{A+B+C+D}{4}={90}^{\circ }$
$⟹\frac{A+C}{4}={90}^{\circ }-\frac{B+D}{4}$
$⟹tan\left(\frac{A+C}{4}\right)=tan({90}^{\circ }-\frac{B+D}{4})$
Since, $cot\theta =tan({90}^{\circ }-\theta )$
$⟹tan\left(\frac{A+C}{4}\right)=cot\left(\frac{B+D}{4}\right)$