Question

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

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

Answer 1

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

$2s=a+b+c$

We know,

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

Similarly,

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

Now,

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

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

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

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

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