Question

In $△ABC,$ if $a,b,c$ are in A.P., then $(\tan \frac{A}{2}+\tan \frac{C}{2})\tan \frac{B}{2}=$

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $\frac{3}{4}$

Answer 1

Answer: (B)

$\frac{2}{3}$

$a,b,c$ are in A.P$\Rightarrow 2b=a+c$
We know that $\tan \frac{A}{2}=\frac{(s-b)(s-c)}{△},$ $\tan \frac{B}{2}=\frac{(s-a)(s-c)}{△},$ $\tan \frac{C}{2}=\frac{(s-a)(s-b)}{△}$
$\therefore (\tan \frac{A}{2}+\tan \frac{C}{2})\tan \frac{B}{2}$
$=[\frac{(s-b)(s-c)}{△}+\frac{(s-a)(s-b)}{△}].\frac{(s-a)(s-c)}{△}$
Multiply and divide by $s(s-b)$ we get
$=[\frac{(s-b)(s-c)}{△}+\frac{(s-a)(s-b)}{△}].\frac{(s-a)(s-c)}{△}\times \frac{s(s-b)}{s(s-b)}$
$=[\frac{(s-b)(s-c)}{△}+\frac{(s-a)(s-b)}{△}].\frac{{△}^2}{△s(s-b)}$ where ${△}^2=s(s-a)(s-b)(s-c)$
$=[\frac{(s-b)(s-c)}{△}+\frac{(s-a)(s-b)}{△}].\frac{△}{s(s-b)}$
$=\frac{s-c+s-a}{s}$ on simplification
$=\frac{2s-a-c}{s}$
$=\frac{2s-(2s-b)}{s}$ where $2s=a+b+c$ or $b+c=2s-a$
$=\frac{b}{s}$ on simplification
$=\frac{2b}{2s}$ by multiplying the numerator and denominator by $2$
$=\frac{2b}{a+b+c}$ where the perimeter $2s=a+b+c$
$=\frac{2b}{2b+b}=\frac{2}{3}$ since $a+c=2b$