Question

In $∆ABC,(a+b+c)\left(\tan \left(\frac{A}{2}\right)+\tan \left(\frac{B}{2}\right)\right)$ is equal to

• A. $2ccot\left(\frac{C}{2}\right)$
• B. $2acot\left(\frac{A}{2}\right)$
• C. $\tan \left(\frac{C}{2}\right)$
• D. None of these

Answer 1

The correct option is A

$2ccot\left(\frac{C}{2}\right)$

Explanation for the correct option:

Solving the expression:

From sine rule, we know in $∆ABC$,

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

Also, we know $s=\frac{a+b+c}{2}$

$$\begin{gathered} \therefore (a+b+c)\left(\tan \left(\frac{A}{2}\right)+\tan \left(\frac{B}{2}\right)\right)=2s\left[\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}+\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}\right] \\ =2s\sqrt{\frac{s-c}{s}}\left[\sqrt{\frac{(s-b)}{(s-a)}}+\sqrt{\frac{(s-a)}{(s-b)}}\right] \\ =2\sqrt{s(s-c)}\left[\frac{s-b+s-a}{\sqrt{(s-a)(s-b)}}\right]\left[\because 2s=a+b+c\right] \\ =2\sqrt{s(s-c)}\left[\frac{c}{\sqrt{(s-a)(s-b)}}\right]\left[\therefore 2s-b-a=c\right] \\ =2c\left[\frac{\sqrt{s(s-c)}}{\sqrt{(s-a)(s-b)}}\right] \\ =2c\text{cot}\frac{C}{2}\mathbf{}\left[\because \text{cot}\frac{C}{2}=\frac{\sqrt{s(s-c)}}{\sqrt{(s-a)(s-b)}}\right] \end{gathered}$$$∆ABC,(a+b+c)\left(\tan \left(\frac{A}{2}\right)+\tan \left(\frac{B}{2}\right)\right)$$=2c\text{cot}\left(\frac{C}{2}\right)$

Hence, option(A) is the correct answer.