Question

In $\Delta ABC$, if $\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}$, then ${\tan }^2\left(\frac{A}{2}\right)+{\tan }^2\left(\frac{C}{2}\right)=$

• A. $\frac{290}{429}$
• B. $\frac{290}{143}$
• C. $\frac{143}{33}$
• D. $\frac{113}{33}$

Answer 1

Answer: (A)

Given that, $\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}=k$
$s-a=11k$
$s-b=12k$
$s-c=13k$
On adding,
$3s-(a+b+c)=36k$
$3s-2s=36k$
[ as $a+b+c=2s]$
$s=36k$
Now, ${\tan }^2\frac{A}{2}+{\tan }^2\frac{C}{2}$
$=\frac{(s-b)(s-c)}{s(s-a)}+\frac{(s-a)(s-b)}{s(s-c)}$
$=\frac{(12k)(13k)}{(36k)(11k)}+\frac{(11k)(12k)}{(36k)(13k)}=\frac{12}{36}\left[\frac{13}{11}+\frac{11}{13}\right]$
$=\frac{1}{3}\cdot \frac{169+121}{11\times 13}]$
$=\frac{290}{429}$