Question

In $△ABC$ , $(a+b+c)$ $(\tan \frac{A}{2}+\tan \frac{B}{2})=$

• A. 2 $c$ cot $\frac{A}{2}$
• B. 2 $c$ cot $\frac{B}{2}$
• C. 2 $c$ cot $\frac{C}{2}$
• D. 2 $c$ tan $\frac{C}{2}$

Answer 1

The correct option is C 2 $c$ cot $\frac{C}{2}$

We have a formula,

$\tan \left(\frac{A}{2}\right)=\sqrt{\frac{(s-c)(s-b)}{s(s-a)}}$

$\tan \left(\frac{B}{2}\right)=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$

Given,

$(a+b+c)(\tan \frac{A}{2}+\tan \frac{B}{2})$

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

solving the above equation, we get,

$=2s\sqrt{\frac{(s-a)(s-c)(s-b+s-a)}{(s-a)(s-b)}}$

$=2c\sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$

$=2c\cot \left(\frac{C}{2}\right)$