Question

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