Question

In a triangle, $ABC,1-\tan \left(\frac{A}{2}\right)\tan \left(\frac{B}{2}\right)$ is equal to

• A. $\frac{2c}{a+b+c}$
• B. $\frac{2c}{a+b-c}$
• C. $\frac{2b}{a+b+c}$
• D. None of these

Answer 1

The correct option is A

$\frac{2c}{a+b+c}$

Explanation for the correct option:

Find the value of the expression:

By using the half angle formula in sides form,

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

$\tan \left(\frac{B}{2}\right)=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$ , where $s=$semi-perimeter and $a,b,c$ are the sides of triangle.

Now, multiply these two terms, we get

$\begin{array}{rcl}\tan \left(\frac{A}{2}\right)\tan \left(\frac{B}{2}\right)& =& \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\times \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}\\ & =& \sqrt{\frac{(s-b)(s-c)(s-a)(s-c)}{s(s-a)\times s(s-b)}}\\ & =& \frac{s-c}{s}\\ & =& 1-\frac{c}{s}\end{array}$

Now, rearrange the term according to finding expression,

$\begin{array}{rcl}1-\tan \left(\frac{A}{2}\right)\tan \left(\frac{B}{2}\right)& =& \frac{c}{s}\\ & =& \frac{2c}{a+b+c}\left[\because s=\frac{a+b+c}{2}\right]\end{array}$

Hence, the correct option is A.