Question

In $\Delta ABC$ if $\tan \frac{A}{2}\tan \frac{B}{2}+\tan \frac{B}{2}\tan \frac{C}{2}=\frac{2}{3}$

then $a+c?$

• A. $3b$
• B. $2b$
• C. $\frac{3b}{2}$
• D. $4b$

Answer 1

The correct option is B $2b$

$tan\frac{A}{2}=\frac{\Delta }{s(s-a)},tan\frac{B}{2}=\frac{\Delta }{s(s-a)},tan\frac{C}{2}=\frac{\Delta }{s(s-c)}$

$\frac{\Delta }{s(s-a)}\frac{\Delta }{s(s-b)}+\frac{\Delta }{s(s-c)}$

$\frac{\Delta }{s(s-b)}=\frac{2}{3}$

$\frac{{\Delta }^2}{s^2(s-b)}\left(\frac{1}{(s-a)+\frac{1}{(s-c)}}\right)=\frac{2}{3}$

$\frac{{\Delta }^2(25-a-c)}{s^2(s-b)(s-c)(s-a)}=\frac{2}{3}$

$\left(\frac{25-a-c}{s}\right)=\frac{2}{3}$

$\frac{2\times (a+b+c-a-c)}{a+b+c}=\frac{2}{3}$

$\frac{b}{b+(a+c)}=\frac{1}{3}$

$(a+c)=2b$