Question

In a cyclic quadrilateral $ABCD;a,b,c,d$ being the lengths of sides $AB,BC,CD,DA$ respectively and $s$ is semi-perimeter of quadrilateral, then the value of ${\tan }^2\frac{B}{2}$ is equal to

• A. $\frac{(s-a)(s-b)}{(s-c)(s-d)}$
• B. $\frac{(s-a)(s-c)}{(s-b)(s-d)}$
• C. $\frac{(s-a)(s-d)}{(s-b)(s-c)}$
• D. none of these

Answer 1

The correct option is A $\frac{(s-a)(s-b)}{(s-c)(s-d)}$

from $△ABC$,we get
${AC}^2=a^2+b^2-2ab\cos B\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(i\right)$
and from $△ADC$
we have, ${AC}^2=c^2+a^2-2cd\cos D$
$=c^2+d^2-2cd\cos ({180}^{\circ }-B)$
${AC}^2=c^2+a^2+2cd\cos B\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(ii\right)$
from $\left(i\right)$ and (ii)
$\cos B=\frac{a^2+b^2-c^2-d^2}{2(ab+cd)}$
Now since,${\tan }^2\frac{B}{2}=\frac{1-\cos B}{1+\cos B}$
${\tan }^2\frac{B}{2}=\frac{2(ab+cd)-(a^2+b^2-c^2-d^2)}{2(ab+cd)+(a^2+b^2-c^2-d^2)}$
$=\frac{{(c+d)}^2-{(a-b)}^2}{{(a+b)}^2-{(c-d)}^2}$
$=\frac{(c+d+a-b)(c+d-a+b)}{(a+b-c+d)(a+b+c-d)}$
$=\frac{(2s-2b)(2s-2a)}{(2s-2c)(2s-2d)}=\frac{(s-a)(s-b)}{(s-a)(s-c)}$ $[sincea+b+c+d=2s]$
${\tan }^2\frac{B}{2}=\frac{(s-a)(s-b)}{(s-d)(s-c)}$