Question

In a $\Delta ABC$ the sides $a,b$ and $c$ are in A.P. Evaluate $(\tan \frac{A}{2}+\tan \frac{C}{2}):\cot \frac{B}{2}$

• A. $2:3$
• B. $1:3$
• C. $3:2$
• D. None of the above
  

Answer 1

The correct option is A $2:3$

$a,b,c$ are in A.P
$\Rightarrow 2b=a+c$
$\Rightarrow 2s=a+b+c=3b$ -----(1)
$\frac{\tan \frac{A}{2}+\tan \frac{C}{2}}{\cot \frac{B}{2}}=\frac{\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}+\frac{\sin \frac{C}{2}}{\cos \frac{C}{2}}}{\frac{\cos \frac{B}{2}}{\sin \frac{B}{2}}}=\frac{\sin \frac{B}{2}}{\cos \frac{A}{2}\cos \frac{C}{2}}$
$=\frac{\sqrt{\frac{(s-a)(s-c)}{ac}}}{\sqrt{\frac{s(s-a)}{bc}}\sqrt{\frac{s(s-c)}{ab}}}=\frac{b}{s}=\frac{2}{3}$ from (1)
$\therefore (\tan \frac{A}{2}+\tan \frac{C}{2}):\cot \frac{B}{2}=2:3$