Question

In any triangle $ABC,1-\tan \left(\frac{A}{2}\right)\tan \left(\frac{B}{2}\right)=$

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

Answer 1

Answer: (C)

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

As we know by using the half angle formula in sides form, The value of $tan\left(\frac{A}{2}\right)$ $=$$\sqrt{\frac{(s-b)(s-c)}{\left(s\right)(s-a)}}$

As we know by using the half angle formula in sides form,The Formula of $tan\left(\frac{B}{2}\right)$ $=$$\sqrt{\frac{(s-a)(s-c)}{\left(s\right)(s-b)}}$

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

$\therefore$ $tan\left(\frac{A}{2}\right)$ $tan\left(\frac{B}{2}\right)$$=$ $\frac{s-c}{s}$

$\therefore$ $1-$ $tan\left(\frac{A}{2}\right)$ $tan\left(\frac{B}{2}\right)$$=$ $\frac{c}{s}$

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

$\therefore$ $1-$ $tan\left(\frac{A}{2}\right)$ $tan\left(\frac{B}{2}\right)$$=$ $\frac{2c}{a+b+c}$

So Option C is the required answer.