Question

If $A+B+C=180°$, $\sum \tan \frac{A}{2}\tan \frac{B}{2}$ is

• A. $0$
• B. $1$
• C. $2$
• D. $3$
• E. $4$

Answer 1

The correct option is B

$1$

Explanation for correct option:

Given,$A+B+C=180°$

Now,

$\begin{array}{rcl}A+B+C& =& 180°\\ \frac{A}{2}+\frac{B}{2}+\frac{C}{2}& =& 90°\\ \frac{A}{2}+\frac{B}{2}& =& 90°-\frac{C}{2}\\ & & Taking"\tan "bothside\\ \tan \left(\frac{A}{2}+\frac{B}{2}\right)& =& \tan \left(90°-\frac{C}{2}\right)\\ \frac{\left[\tan \frac{A}{2}+\tan \frac{B}{2}\right]}{\left[1–\tan \frac{A}{2}\tan \frac{B}{2}\right]}& =& cot\frac{C}{2}\\ \frac{\left[\tan \frac{A}{2}+\tan \frac{B}{2}\right]}{\left[1–\tan \frac{A}{2}\tan \frac{B}{2}\right]}& =& \frac{1}{\tan {\displaystyle \frac{C}{2}}}\\ \left[\tan \frac{A}{2}+\tan \frac{B}{2}\right]\times \tan \frac{C}{2}& =& \left[1–\tan \frac{A}{2}\tan \frac{B}{2}\right]\\ \tan \frac{A}{2}\tan \frac{C}{2}+\tan \frac{B}{2}\tan \frac{C}{2}+\tan \frac{A}{2}\tan \frac{B}{2}& =& 1\\ & & \\ & & \end{array}$

Thus, $\sum \tan \frac{A}{2}\tan \frac{B}{2}=1$

Hence, correct option is $\left(B\right)$