Question

In a triangle $ABC,$ if $a,b,c$ are in A.P., then $\tan \frac{A}{2}\tan \frac{C}{2}=$

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

Answer 1

The correct option is A $1$

Since $a,b,c$ are in A.P $2b=a+c$
We know the half-angle formula
$\tan \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}},$ $\tan \frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$
Now,$\tan \frac{A}{2}\tan \frac{C}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$
$=\frac{(s-b)}{s}$ on simplification
Multiply both sides by $2$ we get
$=\frac{(2s-2b)}{2s}$
$=\frac{(a+b+c-2b)}{a+b+c}$ where $2s=a+b+c$
$=\frac{a+c-b}{a+b+c}$
Since $a,b,c$ are in A.P we have $a+c=2b$
$=\frac{2b-b}{2b+b}=\frac{1}{3}$ on simplification